mst124 Unit8 E1i1 PDF

Unit 8
Integration methods
Introduction
Introduction
In this unit you’ll continue your study of integration. In Unit 7 you saw
that integration is the reverse of differentiation. You saw that you can use
it to solve problems in which you know the values taken by the rate of
change of a continuously changing quantity, and you want to work out the
values taken by the quantity, or how much the quantity changes over some
period. For example, you saw that if you have a formula for the velocity of
an object in terms of time, then you can use integration to find a formula
for the displacement of the object in terms of time, or work out the change
in the displacement of the object over some period of time.
In Sections 1 and 2 of this unit, you’ll meet a different way to think about
integration. This way of thinking about it involves areas of regions of the
plane, rather than rates of change. You’ll see that the link between the two
ways of thinking about integration helps you to solve problems both about
areas and about rates of change.
In Sections 3 to 5 of the unit, you’ll learn some more methods for
integrating functions, which you can use in addition to the methods that
you learned in Unit 7. These further methods will allow you to integrate a
much wider variety of functions than you’ve learned to integrate so far.
Many students find some of the material in this unit quite challenging
when they first meet it, so don’t worry if that’s the case for you too. As
with most mathematics, it will seem more straightforward once you’re
more familiar with it. If you find a particular technique difficult, then
make sure that you watch the associated tutorial clips, and use the
practice quiz and exercise booklet for this unit for more practice.
This unit is likely to take you more time to study than most of the other
units, so the study planner allows extra time for it.
105
Unit 8 Integration methods
1 Areas, signed areas and definite

integrals
In this section you’ll look at a type of mathematical problem that you
might think has little connection with the mathematics of rates of change
that you studied in Units 6 and 7. In Section 2 you’ll see that in fact the
two areas of mathematics are closely related.
Throughout Sections 1 and 2, we’ll work with continuous functions.
Remember that a continuous function is a function whose graph has no
breaks in it, over its whole domain. Informally, it’s a function whose graph
you can draw without taking your pen off the paper (though you might
need an infinitely large piece of paper, as the graph might be infinitely
long!).
1.1 Areas and signed areas

Sometimes it’s useful to calculate the area of a flat shape with a curved
boundary. For example, suppose that an architect is designing a large
building shaped as shown in Figure 1(a). The curve of the roof is given by
the graph of the function
1 2
f (x) = 9 − 36 x (−18 ≤ x ≤ 18),
where x and f (x) are measured in metres. This graph is shown in
Figure 1(b).
Figure 1 (a) The shape of a building (b) the graph that gives the curve
of the roof
106
1 Areas, signed areas and definite integrals
Suppose that the architect wants to calculate the area shaded in

Figure 1(b), so that she can determine the amounts of materials needed to
build the end wall of the building.
Here’s a way of calculating an approximate value for this area, which you
can make as accurate as you wish. You start by dividing the interval
[−18, 18], which is the interval of x-values that corresponds to the curve,
into a number of subintervals of equal width.
For example, in Figure 2 the interval has been divided into twelve
subintervals. The right endpoint of the first subinterval is equal to the left
endpoint of the next subinterval, and so on.
For each subinterval, you approximate the shape of the curve on that
subinterval by a horizontal line segment whose height is the value of the
function at the left endpoint of the subinterval, as shown in Figure 2. You
calculate the areas of the rectangles between these line segments and the
x-axis (simply by multiplying their heights by their widths in the usual
way), and add up all these areas to give you an approximate value for the
required area. The more subintervals you use, the better will be the
approximation.
Figure 2 A collection of rectangles whose total area is approximately the

area in Figure 1(b)
In the next example this method is used, with four subintervals, to find an
approximate value for the area discussed above. Notice that, in both the
example and in Figure 2 above, the rectangle corresponding to the first
subinterval has zero height and therefore zero area – this happens because
the value of the function at the left endpoint of this subinterval is zero.
107
Example 1 Calculating an approximate value for an area

Use the method described above, with four subintervals as shown
below, to find an approximate value for the area between the graph of
1 2
the function f (x) = 9 − 36 x and the x-axis.
Solution
We divide the interval [−18, 18] into four subintervals of equal width.
The whole interval has width 36, so each subinterval has width
36/4 = 9.
The left endpoints of the four subintervals are
−18, −18 + 9, −18 + 2 × 9, −18 + 3 × 9,
that is
−18, −9, 0, 9.
So the heights of the rectangles are
f (−18), f (−9), f (0), f (9),
that is,
9− 1
36 × (−18)2 , 9− 1
36 × (−9)2 , 9− 1
36 × 02 , 9− 1
36 × 92 ,
which evaluate to
27 27
0, 4 , 9, 4 .
So we obtain the following approximate value for the area:
405
(0 × 9) + ( 27 27
4 × 9) + (9 × 9) + ( 4 × 9) = 2 = 202.5.
The calculation in Example 1 shows that a (rather crude) approximate

value for the area of the cross-section of the roof discussed at the
beginning of this subsection is 202.5 m2 .
108
Note that the units for the area of a region on a graph are, as you’d
expect, the units on the vertical axis times the units on the horizontal axis.
For example, if the units on both axes are metres, then the units for area
are square metres (m2 ). If a graph has no specific units on the axes, then
the units for area are simply ‘square units’, and we usually omit them
when we state an area.
In the next activity you’re asked to find another approximation for the
area of the cross-section of the roof discussed earlier, by using the
subinterval method with six subintervals instead of four.
Activity 1 Calculating an approximate value for an area
Use the method described above, with six subintervals as shown below, to
find an approximate value for the area between the graph of the function
1 2
f (x) = 9 − 36 x and the x-axis.
Table 1 shows the approximate values for the area between the graph of
1 2
the function f (x) = 9 − 36 x and the x-axis that were found in Example 1
and in the solution to Activity 1. It also shows, to three decimal places,
some further approximate values that were found in the same way, but
using larger numbers of subintervals. The calculations were carried out
using a computer.
Table 1 Approximate values for the area in Figure 1(b)

Number of subintervals 4 6 12 50 100 500
Approximation obtained 202.5 210 214.5 215.914 215.978 215.999
You can see that as the number of subintervals gets larger and larger, the
approximation obtained seems to be getting closer and closer to a
particular number. We say that this number is the limit of the
approximations as the number of subintervals tends to infinity. It’s the
1 2
exact value of the area between the graph of the function f (x) = 9 − 36 x
and the x-axis. From Table 1, it looks as if the exact value of this area is
216, or perhaps a number very close to 216.
109
In general, if you have a continuous function f whose graph lies on or

above the x-axis throughout an interval [a, b], then you can use the method
demonstrated above to find, as accurately as you want, the area between
the graph of f and the x-axis, from x = a to x = b. Figure 3(a) illustrates
an area of this type, and Figure 3(b) illustrates the approximation to the
area that’s obtained by using 8 subintervals.
Figure 3 (a) The area between a graph and the x-axis, from x = a to
x = b (b) A collection of rectangles whose total area is approximately
this area
Here’s a summary of the method that you’ve met for finding an
approximate value for an area of this type. You start by dividing the
interval [a, b] into a number, say n, of subintervals of equal width. The
width of each subinterval is then (b − a)/n. For each subinterval you
calculate the product
, -
left endpoint b−a
f × , (1)
of subinterval n
and you add up all these products. In general, the larger the number n of
subintervals, the closer your answer will be to the area between the graph
of f and the x-axis, from x = a to x = b.
Now suppose that you have a continuous function f whose graph lies on or
below the x-axis throughout an interval [a, b], and you want to calculate
the area between the graph of f and the x-axis, from x = a to x = b, as
illustrated in Figure 4(a).
You can use the method above, with a small adjustment at the end, to
calculate an approximate value for an area like this. Consider what
happens when you apply the method to a graph like the one in
Figure 4(a). You start by dividing the interval [a, b] into n subintervals of
equal width, then you calculate all the products of form (1) and add them
all up. For a graph like this, because the value of f at the left endpoint of
each subinterval is negative (or possibly zero), each product of form (1)
will also be negative (or zero). In fact, each product of form (1) will be the
negative of the area between the line segment that approximates the curve
and the x-axis, as illustrated in Figure 4(b). So when you add up all the
products of form (1), you’ll obtain an approximate value for the negative of
the area between the curve and the x-axis, from x = a to x = b.
110
Figure 4 (a) The area between another graph and the x-axis, from x = a
to x = b (b) A collection of rectangles whose total area is approximately
this area
That’s not a problem, because you can simply remove the minus sign to
obtain the approximate value for the area that you want. However, to help
us deal with situations like this, it’s useful to make the following
definitions. These definitions will be important throughout the unit.
Consider any region on a graph that lies either entirely above or entirely
below the x-axis. The signed area of the region is its area with a plus or
minus sign according to whether it lies above or below the x-axis,
respectively. For example, in Figure 5 the two shaded regions above the
x-axis have signed areas +4 and +6, respectively, which you can write
simply as 4 and 6, and the shaded region below the x-axis has signed
area −3.
Figure 5 Regions on a graph

If you have a collection of regions on a graph, where each region in the
collection lies either entirely above or entirely below the x-axis, then the
total signed area of the collection is the sum of the signed areas of the
individual regions. For example, the total signed area of the collection of
three regions in Figure 5 is 4 + (−3) + 6 = 7.
The units for signed area on a graph are, as you’d expect, the same as the
units for area on the graph. That is, they’re the units on the vertical axis
times the units on the horizontal axis. If there are no specific units on the
axes, then we don’t use any specific units for signed area.
111
Example 2 Understanding signed areas

The areas of some regions on a graph are marked below.
In each of parts (a)–(c), use these areas to find the signed area
between the graph and the x-axis, from the first value of x to the
second value of x.
(a) From x = −3 to x = −2. (b) From x = −2 to x = 1.
(c) From x = −3 to x = 1.
Positive signed area
Negative signed area Solution

(a) The signed area from x = −3 to x = −2 is −2.2.
(b) The signed area from x = −2 to x = 1 is +4.8 = 4.8.
(c) The signed area from x = −3 to x = 1 is −2.2 + 4.8 = 2.6.
Of course, the signed area value found in Example 2(c) doesn’t correspond
to any actual area on the graph, as it’s the sum of a positive signed area
and a negative signed area.
112
Activity 2 Understanding signed areas
Consider again the graph in Example 2. In each of parts (a)–(d) below, use
the given areas to find the signed area between the graph and the x-axis,
from the first value of x to the second value of x.
(a) From x = 1 to x = 5. (b) From x = 5 to x = 7.
(c) From x = 1 to x = 7. (d) From x = 1 to x = 9.
With the definition of signed area that you’ve now seen, the method that
you’ve met in this subsection can be described concisely as follows.
Strategy:
To find an approximate value for the signed area between
the graph of a continuous function f and the x-axis, from
x = a to x = b
Divide the interval between a and b into n subintervals, each of
width (b − a)/n. For each subinterval, calculate the product
, -
endpoint of subinterval b−a
f × ,
nearest a n
and add up all these products.
In general, the larger the number n of subintervals, the closer your
answer will be to the required signed area.
Notice that the box above uses the phrase ‘endpoint of subinterval
nearest a’, rather than ‘left endpoint of subinterval’, which means the same
thing. This will be convenient later in this subsection. (Note that, in this
phrase, it’s the endpoint of the subinterval, not the subinterval itself,
that’s nearest a!)
Figures 6 and 7 illustrate the strategy above. If you add up the signed
areas of all the rectangles in Figure 6(a), then you’ll obtain an
approximate value for the signed area in Figure 6(b).
Figure 6 (a) A collection of rectangles whose total signed area is

approximately the signed area shown in (b)
113
Similarly, if you add up the signed areas of all the rectangles in Figure 7(a),
then you’ll obtain an approximate value for the signed area in Figure 7(b).
Figure 7 (a) Another instance of a collection of rectangles whose total

signed area is approximately the signed area shown in (b)
Activity 3 Calculating an approximate value for a signed area
Use the method described in the box above, with six subintervals as shown
on the left below, to find an approximate value for the signed area between
the graph of the function f (x) = 3 − x2 and the x-axis from x = −3 to
x = 3, as shown on the right below.
In the next activity you can use a computer to explore approximations to

signed areas found by using subintervals in the way that you’ve seen.
114
Activity 4 Calculating more approximate values for signed areas
Open the applet Approximations for signed areas. Check that the function
is set to f (x) = 3 − x2 , the values of a and b are set to −3 and 3
respectively, and the number of subintervals is set to 6. The resulting
approximation for the signed area between the graph of f and the x-axis,
from x = a to x = b, should then be the value found in the solution to
Activity 3.
Now increase the number of subintervals, and observe the effect on the
approximation. What do you think is the exact value of the signed area?
Experiment by changing the function, the values of a and b, and the
number of subintervals, to find approximate values for some other signed
areas.
So far in this subsection, whenever a signed area from x = a to x = b has

been discussed, the value of b has been greater than the value of a.
The value of b can also be equal to the value of a, and this gives a signed
area of zero, as illustrated in Figure 8. In cases like this, the method that
you’ve seen for using subintervals to find approximate values for signed
areas gives the answer zero, since the width (b − a)/n of each subinterval is
zero.
Figure 8 A signed area equal to zero

In fact, for reasons that you’ll see later in the unit, it’s useful to extend the
definition of signed area a little, to give a meaning to the phrase ‘the
signed area between the graph of f and the x-axis from x = a to x = b’
when b is less than a. It’s not immediately obvious what the phrase means
in this case, but it turns out that the natural meaning is as follows. This
definition will be important throughout the unit.
115
Suppose that f is a continuous function whose domain includes the

interval [b, a], as illustrated in Figure 9. Then the signed area between
the graph of f and the x-axis from x = a to x = b is defined to be the
negative of the signed area between the graph of f and the x-axis from
x = b to x = a.
Figure 9 The graph of a continuous function f whose domain includes

the interval [b, a]
For example, consider the graph in Figure 10. The signed area from x = 1
to x = 3 is 4, so the signed area from x = 3 to x = 1 is −4. Similarly, the
signed area from x = 6 to x = 8 is −5, so the signed area from x = 8
to x = 6 is −(−5) = 5.
Figure 10 Areas on a graph

Because of this extended definition of signed area, you now have to think a
little more carefully when you read or use the phrase ‘signed area’ (and the
signed area isn’t zero). If there’s no mention of ‘from x = a to x = b’, then
to decide whether the signed area is positive or negative you just need to
consider whether it lies above or below the x-axis. However, if the text is
of the form ‘the signed area from x = a to x = b’, then to decide whether
the signed area is positive or negative you also have to consider whether b
is greater than or less than a.
This is illustrated in the next example. A helpful way to think about
whether b is greater than or less than a is to think about whether x moves
forward or backward as it changes from x = a to x = b.
116
Example 3 Understanding the extended definition of signed area

The areas of some regions on a graph are marked below. In each of
parts (a)–(d), use the given areas to find the signed area between the
graph and the x-axis, from the first value of x to the second value
of x.
(c) From x = 10 to x = 6. (d) From x = 6 to x = −1.
Solution
(a) From 3 to 3 is from a number to the same number.
The signed area from x = 3 to x = 3 is 0.
(b) From 6 to 10 is forward.
The signed area from x = 6 to x = 10 is −5.4.
(c) From 10 to 6 is backward.
The signed area from x = 6 to x = 10 is −5.4,
so the signed area from x = 10 to x = 6 is 5.4.
(d) From 6 to −1 is backward.
The signed area from x = −1 to x = 6 is −2.4 + 4.0 = 1.6,
so the signed area from x = 6 to x = −1 is −1.6.
In each part of the next activity, start by thinking about whether x moves
forward or backward (or stays fixed) as it moves from the first value of x
to the second value of x.
117
Activity 5 Understanding the extended definition of signed area
The graph in Example 3 is repeated below. In each of parts (a)–(g), use

the given areas to find the signed area between the graph and the x-axis,
from the first value of x to the second value of x.
(c) From x = 2 to x = 10. (d) From x = 10 to x = 2.
(e) From x = 8 to x = 8. (f) From x = 2 to x = −3.
(g) From x = 10 to x = −1.
The subinterval method for finding approximate values for signed areas,
which is summarised in the box on page 113, applies no matter whether
the value of b is greater than, less than, or equal to the value of a. The
reason why it works in cases where b is less than a is that in these cases
the ‘width’ (b − a)/n of each subinterval is negative.
You might like to use the applet Approximations for signed areas to see
examples of the method giving approximate values for signed areas in cases
where b is less than a.
Of course, if you want to work out an area between a curve and the x-axis
over an interval [a, b] by using the method that you’ve seen in this
subsection, then normally you would work out the signed area from a to b,
rather than the signed area from b to a. However, it’s important to
understand what’s meant by the signed area from b to a, for reasons that
you’ll see later in the unit.
118
1.2 Definite integrals

Before you learn more about why signed areas are important, it’s useful for
you to learn some terminology and notation that are used when working
with them.
If f is a continuous function and a and b are numbers in its domain, then
the signed area between the graph of f and the x-axis from x = a to x = b
is called the definite integral of f from a to b, and is denoted by
! b
f (x) dx.
a
This notation is read as ‘the integral from a to b of f of x, d x’. The
numbers a and b are called the lower and upper limits of integration,
respectively.
For example, for the function f in Figure 11,
! 7 ! 9
f (x) dx = −9, f (x) dx = 2,
3 7
! 9
and f (x) dx = −9 + 2 = −7.
3
Similarly, for the same function,
! 4 ! 3
f (x) dx = 0, and f (x) dx = −(−9) = 9.
4 7
Figure 11 The graph of a function, and some areas

You’ll see a little later in this subsection where this notation for a definite
integral comes from.
!
The symbol is called the integral sign. It sometimes appears in a
/
smaller form, , when it’s in a line of typed text.
119
As with Leibniz notation for derivatives, the ‘d’ in the notation for definite
integrals has no independent meaning. In many texts, including this one,
it’s printed in upright type, rather than italic type, to emphasise that it’s
not a variable. You don’t need to do anything special when you
handwrite it.
/b
When you use the notation a f (x) dx, remember that you must include
/b
not only the a at the beginning, but also the dx at the end. Try not to
forget the dx!
You’ve now seen that a definite integral is a type of signed area, and you
saw in Unit 7 that an indefinite integral is a type of general antiderivative.
These two seemingly unrelated concepts are, as you probably suspect,
closely related. You’ll see how in Section 2.
The box below lists some standard properties of definite integrals, which
come from their definition as signed areas. These properties hold for all
values of a, b and c in the domain of the continuous function f . The
second property comes from the extended definition of signed area, to
cases where b is less than a. The third property expresses the fact that the
signed area from a to c is equal to the signed area from a to b plus the
signed area from b to c. Figure 12 illustrates this property in a case where
a < b < c, but the extended definition of signed area ensures that the
property holds even when a, b and c aren’t in this order. This is one reason
why the definition of signed area is extended in the way that you’ve seen.
Standard properties of definite integrals

! a
f (x) dx = 0
a
! a ! b
f (x) dx = − f (x) dx
b a
! c ! b ! c
f (x) dx = f (x) dx + f (x) dx
a a b
Figure 12 Two adjacent signed areas
120
Activity 6 Understanding definite integrals
Consider the function f whose graph is shown below. The areas of some
regions are marked. By using these areas, write down the values of the
following definite integrals.
Hint: notice that in some of these definite integrals the upper limit of
integration is less than the lower limit of integration.
! −3 ! 2 ! 7
(a) f (x) dx (b) f (x) dx (c) f (x) dx
−5 −3 2
! 2 ! 7 ! 7
(d) f (x) dx (e) f (x) dx (f) f (x) dx
−5 −3 −5
! 5 ! −3 ! 2
(g) f (x) dx (h) f (x) dx (i) f (x) dx
5 2 7
! −5 ! −3 ! −5
(j) f (x) dx (k) f (x) dx (l) f (x) dx
−3 7 7
Activity 7 Using a standard property of definite integrals
Consider again the graph in Activity 6.

! 9 ! 9
Given that f (x) dx = −15, find f (x) dx.
2 7
121
As you’d expect, you can replace the expression f (x) in the notation for a
definite integral by the formula for a particular function. For example, the
signed areas in Figure 13 are denoted by
! 1 ! 1
x2 dx and (x3 − 1) dx,
−1 0
respectively.
Figure 13 Signed areas between the graphs of particular functions and

the x-axis
As always with algebraic notation, the notation for a definite integral can
be used with letters other than the standard ones. For example, if g is a
continuous function whose domain contains the numbers p and q, and you
use t to denote the input variable of g, then the definite integral of g
from t = p to t = q is denoted by
! q
g(t) dt.
p
In fact, the input variable of the function in a definite integral is what’s

known as a dummy variable – you can change its name to any other
variable name that you like, without affecting the value of the definite
integral. For example, if f is a continuous function whose domain contains
the numbers a and b, then
! b ! b ! b
f (x) dx = f (t) dt = f (u) du,
a a a
and so on. As a particular example,
! 1 ! 1 ! 1
2 2
x dx = t dt = u2 du,
−1 −1 −1
since all these definite integrals denote the signed area shown in
Figure 13(a).
If f is any continuous function, and a and b are numbers in its domain,
/b
then you can find an approximate value for the definite integral a f (x) dx,
as accurately as you like, by using the method that you met in the last
subsection. Here’s the method expressed algebraically.
122
/b
Suppose that you want to find a f (x) dx, as illustrated in Figure 14(a).
You divide the interval between a and b into n subintervals, each of
width (b − a)/n, as illustrated in Figure 14(b). We’ll denote (b − a)/n
by w here, for conciseness.
/b
Figure 14 (a) A definite integral a f (x) dx (b) A collection of n
rectangles whose total signed area is approximately this definite integral
The endpoints nearest a of the subintervals are
a + 0w, a + 1w, a + 2w, ..., a + (n − 1)w.
(Remember that the strategy on page 113 specifies ‘endpoint of subinterval

nearest a’, rather than ‘left endpoint of subinterval’. The reason for this is
that it ensures that the algebraic expressions for the endpoints stated
above are correct when b is less than a, as well as when b is greater than a
or equal to a.)
So an approximate value for the definite integral of f from x = a to x = b
is given by
f (a + 0w) × w + f (a + 1w) × w + f (a + 2w) × w + · · ·
· · · + f (a + (n − 1)w) × w.
This expression can be simplified slightly by taking out the common
factor w. Doing this (and writing the common factor at the end) gives
. #
f (a + 0w) + f (a + 1w) + f (a + 2w) + · · · + f (a + (n − 1)w) w.
As the value of n gets larger and larger, the value of this expression gets
/b
closer and closer to the value of a f (x) dx.
The limit of the value of the expression, as the value of n tends to infinity,
/b
is the exact value of a f (x) dx. This idea is summarised below. The
notation ‘ lim ’ means ‘the limit, as n tends to infinity, of’.
n→∞
123
Algebraic definition of a definite integral

Suppose that f is a continuous function and a and b are numbers in
its domain. Then the definite integral of f from x = a to x = b is
given by the equation
! b .
f (x) dx = lim f (a + 0w) + f (a + 1w) + f (a + 2w) + · · ·
a n→∞
#
· · · + f (a + (n − 1)w) w
where w = (b − a)/n.
If f is a particular continuous function, and a and b are numbers in its

/b
domain, then you can sometimes calculate an exact value for a f (x) dx by
using an algebraic method to find the limit in the box above. In some
/b
cases you might even be able to obtain a general formula for a f (x) dx, in
terms of a and b. However, this is usually a complicated and difficult
process, so normally we don’t do it. Instead, if you want to evaluate a
/b
definite integral a f (x) dx, then you have two main options.
The first option is to obtain an approximate value by using subintervals in
the way that you’ve seen. We usually use a computer to carry out the
calculations. The computer algebra system that you’re using in this
module includes a command that instructs the computer to find an
approximate value for a definite integral, to a particular level of precision,
by using a version of the subinterval method that you’ve seen. You’ll learn
about this command in the final subsection of this unit.
/b
The other option for evaluating a definite integral a f (x) dx is to use an
algebraic method that arises from the fundamental theorem of calculus,
which you’ll meet in Section 2. This algebraic method involves much less
calculation than using subintervals, and it has the advantage of giving an
exact answer. However, sometimes it’s not possible or not practicable to
use this method, as you’ll see.
Now here’s an explanation of where the notation and terminology for
definite integrals come from. They arise directly from the method that
you’ve seen for calculating approximate values for definite integrals. To see
how, consider applying the method to a function f , as illustrated in
Figure 15(a).
Consider any one of the subintervals. You could denote its endpoint
nearest a by x, and its width by δx, as shown in Figure 15(b). As you saw
in Unit 6, the notation δx represents a small change in x.
124
Figure 15 (a) The subinterval method applied to a function f from

x = a to x = b (b) One of the subintervals and its corresponding rectangle
With this notation, the signed area of the rectangle corresponding to the
subinterval, from x to x + δx, is
f (x) δx. (2)
So the total signed area of all the rectangles is the sum of a number of
terms, each of form (2). As the number of subintervals gets larger and
larger, the size of δx gets smaller and smaller, and the total signed area of
the rectangles gets closer and closer to the definite integral of f from a
to b. So, loosely, you can think of this definite integral as the sum of
infinitely many terms of form (2), where the quantity δx is infinitely small.
Historically, as you saw in Unit 6, the quantity δx was denoted by dx when
it becomes infinitely small, so the sum described above was denoted by
! b
f (x) dx,
a
!
where the symbol is an elongated ‘S’, which stands for ‘sum’.
Johann Bernoulli (1667–1748)
To see where the term ‘integral’ comes from, remember that the verb ‘to
integrate’ means ‘to join together’. Loosely, you can think of a definite
integral as being obtained by joining together infinitely many signed areas,
each infinitely narrow, into a single signed area.
The name and principal symbol for integral calculus were discussed
by Gottfried Wilhelm Leibniz and the Swiss mathematician
Johann Bernoulli.
/ Leibniz preferred ‘calculus summatorius’ as the
name, and , the elongated S, as the symbol. Bernoulli preferred
‘calculus integralis’ as the name, and the capital letter I as the
symbol. Eventually they agreed to compromise, / adopting Bernoulli’s
name ‘integral calculus’ and Leibniz’s symbol . The first appearance
in print of the word ‘integral’ was in a work by Johann’s brother,
Jacob Bernoulli, in 1690, though Johann insisted that he had been
the one to introduce the term.
Jacob Bernoulli (1654–1705)
125
The modern notation for a definite integral, with limits at the bottom
and top of the integral sign, is due to the French mathematician
Joseph Fourier, who introduced it in his pioneering book on heat
diffusion Théorie analytique de la chaleur (The analytic theory of
heat) of 1822. He first published the notation four years previously in
an announcement for the book. Fourier accompanied
Napoleon Bonaparte on his Egyptian expedition of 1798, and later
supported Jean-François Champollion, the decipherer of the Rosetta
Stone. He is remembered today for initiating the investigation of
Fourier series. These are infinite trigonometric series, which are now
named after him, and which arose in the context of his work on heat
diffusion. You’ll meet the idea of a series in Unit 10.
Joseph Fourier (1768–1830)

To finish this section, here’s a summary of the main idea that you’ve met.
Definite integrals
Suppose that f is a continuous function, and a and b are numbers in
its domain. The signed area between the graph of f and the x-axis
from x = a to x = b is called the definite integral of f from a to b,
and is denoted by
! b
f (x) dx.
a
2 The fundamental theorem of

calculus
In this section you’ll discover how the ideas about signed areas that you
met in Section 1 are connected with the ideas about rates of change that
you met in Units 6 and 7. The two areas of mathematics are linked by an
important fact known as the fundamental theorem of calculus.
This theorem is stated in the first subsection of this section, preceded by
an explanation of why it’s true. It’s a good idea to read the explanation,
to increase your understanding, particularly if you intend to study more
mathematics after this module. However, if you find the explanation hard
to understand, then you can skip it, and go straight to the box headed
‘Fundamental theorem of calculus’, near the end of Subsection 2.1. Read
the contents of this box, and continue reading to the end of the subsection.
Make sure that you understand the fundamental theorem of calculus, and
the other important facts stated there, even if you don’t know why they’re
126
2 The fundamental theorem of calculus
true. The two subsections that follow, Subsections 2.2 and 2.3, show you
how to use the fundamental theorem of calculus. You might like to try
reading Subsection 2.1 again later, as some of the ideas that you found
difficult the first time might seem easier once you’re more familiar with the
ways in which you can use the fundamental theorem of calculus.
2.1 What is the fundamental theorem of calculus?

The first step in obtaining a link between signed areas and rates of change
is to define, for any continuous function f , a new function whose values are
signed areas between the graph of f and the x-axis. Here’s how we do that.
Consider any continuous function f , as illustrated in Figure 16.
Figure 16 The graph of a continuous function f

Let’s choose any number, say s, in the domain of f , and define a new
function, A, to have the same domain as f and the following rule:
A(x) = signed area between the graph of f
and the x-axis, from s to x.
This definition is illustrated in Figure 17.
Figure 17 A continuous function f and a signed area A(x)

We call the function A the signed-area-so-far function for the
function f , with starting point s. Its rule can be expressed concisely as
follows:
! x
A(x) = f (t) dt.
s
127
(Here the variable t has been used in the definite integral instead of the
usual variable x, to avoid confusion with the input variable x of the
function A.)
Note that if a and b are any two numbers in the domain of f , then the
signed area between the graph of f and the x-axis, from x = a to x = b, is
given by
A(b) − A(a).
In other words,
! b
f (x) dx = A(b) − A(a).
a
This fact is illustrated in Figure 18, in a case where s < a < b. The third
property of definite integrals in the box on page 120 tells you that the fact
holds even when s, a and b aren’t in this order.
/b
Figure 18 An illustration of the fact that a
f (x) dx = A(b) − A(a)
Now let’s consider how the value of a signed-area-so-far function changes
as the value of the input variable x changes. Let’s start by looking at what
happens when the function f is a constant function.
128
For example, consider the constant function f (x) = 3, whose graph is

shown in Figure 19. Let’s take the starting point s for the
signed-area-so-far function A to be 1, as indicated in Figure 19.
Figure 19 The graph of the constant function f (x) = 3
Figure 20 shows some values of A(x).
Figure 20 Some values of A(x) when f is the constant function f (x) = 3

and the starting point s is 1
You can see that the signed area A(x) changes at the rate of 3 square units
for every unit by which x changes. For example,
if x increases by 1, then A(x) increases by 3
if x increases by 0.5, then A(x) increases by 0.5 × 3 = 1.5
if x increases by −1, then A(x) increases by −1 × 3 = −3.
(Remember that a negative increase is a decrease.)
In other words, the rate of change of A(x) with respect to x is 3.
129
More generally, you can see that for any constant function f (x) = k, and
any signed-area-so-far function A for f , the value of A(x) changes at the
rate of k square units for every unit by which x changes. In other words,
the rate of change of A(x) is k.
There are two more examples of this in Figure 21. In Figure 21(a), the
function is f (x) = 1.5, so the signed area A(x) changes at the rate of
1.5 square units for each unit by which x changes. In Figure 21(b), the
function is f (x) = −2, so the signed area A(x) changes at the rate of
−2 square units for each unit by which x changes.
Figure 21 Signed-area-so-far functions for constant functions, with

starting points s = −1 in (a) and s = 3 in (b)
Now let’s look at what happens for a function f that isn’t a constant
function. Consider, for example, the function f whose graph is shown in
Figure 22. The starting point s for the signed-area-so-far function A has
been chosen to be 2.
Figure 22 A signed-area-so-far function for a function f that isn’t a

constant function
For this function, as the value of x changes, the value of A(x) doesn’t
change at a constant rate. For example, as illustrated in Figure 23(a),
when x is close to 5 the value of A(x) changes at the rate of about 3 square
units for every unit by which x increases. Similarly, as illustrated in
130
Figure 23(b), when x is close to 7 the value of A(x) changes at the rate of
about 4 square units for every unit by which x increases.
Figure 23 (a) When x is close to 5, the rate of change of A(x) is about 3

(b) When x is close to 7, the rate of change of A(x) is about 4
More precisely, when x = 5 the instantaneous rate of change of A(x) is 3,
and when x = 7 the instantaneous rate of change of A(x) is 4. In fact –
and this is the crucial thing to realise – at any value of x, the
instantaneous rate of change of A(x) is f (x), as illustrated in Figure 24.
Figure 24 The instantaneous rate of change of A(x) at x is f (x)

In other words, the derivative of the function A is the function f , and so
the function A is an antiderivative of the function f .
You can see that this will be true for any continuous function f and any
signed-area-so-far function A for f . That is, we have the important fact
below. Because of its importance, it deserves to be called a theorem.
Theorem
Suppose that f is a continuous function, and s is any number in its
domain. Let A be the signed-area-so-far function for f with starting
point s. Then A is an antiderivative of f .
It’s this fact that leads to the fundamental theorem of calculus. Consider
any continuous function f . Let s be any number in its domain, and let A
be the signed-area-so-far function for f with starting point s.
131
Suppose that a and b are any numbers in the domain of f . Then, as you
/b
saw earlier in this subsection, the value of a f (x) dx, the signed area
between the graph of f and the x-axis from x = a to x = b, is given by
! b
f (x) dx = A(b) − A(a). (3)
a
Now let F be any antiderivative of f . Then, since A and F are both

antiderivatives of f , the value of F (b) − F (a) is the same as the value of
A(b) − A(a). That’s because, as you saw in Unit 7, the value of
F (b) − F (a) is the same no matter which antiderivative of f you take F
to be.
So you can replace A(b) − A(a) by F (b) − F (a) in equation (3) above.
This gives the crucial fact that’s known as the fundamental theorem of
calculus. It’s stated concisely below.
Fundamental theorem of calculus

Suppose that f is a continuous function whose domain contains the
numbers a and b, and that F is an antiderivative of f . Then
! b
f (x) dx = F (b) − F (a).
a
This theorem can be stated in words as follows.
Fundamental theorem of calculus (in words)

numbers a and b, and that F is an antiderivative of f . Then the
signed area between the graph of f and the x-axis from x = a to
x = b is equal to the change in the value of F as x changes from x = a
to x = b.
The equation in the fundamental theorem of calculus is true no matter

whether the value a is less than, equal to or greater than the value b. This
is the main reason why the definition of signed area is extended in the way
that you saw earlier in this unit.
Here’s another important fact that’s come out of the discussion in this
subsection. The theorem in the box on page 131 tells you immediately that
the following fact is true.
Every continuous function has an antiderivative.
132
In fact there’s some disagreement among mathematicians about exactly

which theorem should be called ‘the fundamental theorem of calculus’.
Everyone agrees that this name should be given to a theorem that links
signed areas with rates of change. However, while some mathematicians
use the name for the theorem that’s given this name in this module, other
mathematicians use it for the theorem in the box on page 131. Still other
mathematicians take the view that these two theorems are two
fundamental theorems of calculus, or two parts of a single fundamental
theorem of calculus. So if you search for ‘fundamental theorem of calculus’
on the internet, or look it up in books, then you’re likely to find a variety
of different results.
2.2 Using the fundamental theorem to find signed

areas
The fundamental theorem of calculus gives you a quick way to find the
signed area between the graph of a continuous function f and the x-axis,
from x = a to x = b, in cases where you know a formula for an
antiderivative F of f . The theorem tells you that this signed area, which is
the definite integral
! b
f (x) dx,
a
is given simply by F (b) − F (a).
For example, suppose that you want to find the area between the graph of
the function f (x) = cos x and the x-axis, from x = 0 to x = π/2, as
illustrated in Figure 25. This area is the definite integral
! π/2
cos x dx.
0
Since an antiderivative of f (x) = cos x is F (x) = sin x (as you saw in
Unit 7), the fundamental theorem of calculus tells you that
! π/2 .π#
cos x dx = sin − sin 0 = 1 − 0 = 1.
0 2
Figure 25 The area between
So the area in Figure 25 is 1 square unit. the graph of f (x) = cos x and
You may find it magical that a complicated area like the one in Figure 25 the x-axis, from x = 0 to
x = π/2
can be found from such a simple calculation. In particular, it may seem
amazing that although the area in Figure 25 depends on infinitely many
values of cos x (namely all the values of cos x for x = 0 to x = π/2), it can
be calculated from the values of an antiderivative of cos x at just two
values of x, namely x = 0 and x = π/2.
Notice that in the calculation above it doesn’t matter which of the
infinitely many antiderivatives of f (x) = cos x you use, because they all
133
differ by a constant and hence they all lead to the same final answer. For
example, using the antiderivative F (x) = sin x + 7 gives
! π/2 . .π# #
cos x dx = sin + 7 − (sin 0 + 7) = 1 + 7 − 0 − 7 = 1.
0 2
It’s usually best to use the simplest antiderivative in calculations of this
kind.
You’ll see another example of the fundamental theorem of calculus used to
find a signed area shortly, and you’ll then have the opportunity to find
some signed areas in this way yourself. First, however, it’s helpful for you
to learn a type of notation that’s convenient in calculations of this kind.
For any function F , the expression
F (b) − F (a)
can be denoted by
) 'b
F (x) a .
For example,
) 'π/2 .π#
sin x 0 = sin − sin 0 = 1 − 0 = 1.
2
You can practise evaluating expressions of this type in the next activity.
Activity 8 Evaluating expressions of the form [F (x)]ba
Evaluate the following expressions. In part (c), give your answer to three
significant figures.
) '5 ) '2π ) '1
(a) 12 x2 3 (b) cos x 0 (c) ex −1
With the square bracket notation introduced above, the fundamental

theorem of calculus can be restated as follows.
Fundamental theorem of calculus (square bracket notation)

numbers a and b, and F is an antiderivative of f . Then
! b
) 'b
f (x) dx = F (x) a .
a
In the next example, the fundamental theorem of calculus is used, together

with the square bracket notation, to find a signed area – that is, to
evaluate a definite integral.
134
Example 4 Evaluating a definite integral

Evaluate the definite integral
! 2
x3 dx.
1
(This definite integral is shown in Figure 26.)
Solution
Use the fundamental theorem of calculus. By the general formula
for the indefinite integral of a power function, an antiderivative of x3
is 14 x4 .
! 2
) '2
x3 dx = 14 x4 1
1
Evaluate this expression.

= ( 14 × 24 ) − ( 14 × 14 )
1 15
=4− 4 = 4 .
Figure 26 The definite
integral in Example 4
You’ve now seen that, when you want to find a signed area, it’s much
quicker and simpler to use the fundamental theorem of calculus than to
use the method involving subintervals that you saw in Subsection 1.1.
Another advantage of using the fundamental theorem of calculus is that it
gives you an exact answer, rather than an approximate one. However,
remember that you can use the fundamental theorem of calculus only when
you can find a formula for the antiderivative that you need. Sometimes it’s
difficult or impossible to find such a formula.
In the next activity, you can try evaluating some definite integrals by using
the fundamental theorem of calculus. Use the square bracket notation, as
in Example 4, since you need to get used to this notation. You can find the
antiderivatives that you need by using the table of standard indefinite
integrals given near the end of Unit 7. This table is also included in the
Handbook.
135
Activity 9 Evaluating definite integrals
Evaluate the following definite integrals (which are shown in Figure 27).
Give exact answers, and (except in parts (a) and (d), where the answer is a
simple fraction or an integer) also give answers to three significant figures.
! 1 ! 2 ! 4
1
(a) x2 dx (b) et dt (c) du
−1 0 1 u
! π/4 ! 1
2 1
(d) sec θ dθ (e) 2
du
−π/4 −1 1 + u
Figure 27 The definite integrals in Activity 9

The next example demonstrates how to evaluate another definite integral
by using the fundamental theorem of calculus. The only difference from
the examples that you’ve seen so far is that the antiderivative that’s
needed for the method can’t be found directly from the table of standard
indefinite integrals. Instead, we have to start by manipulating the given
function to express it as a sum of constant multiples of functions that we
can integrate, then use the constant multiple rule and sum rule for
antiderivatives, together with the table of standard indefinite integrals.
You practised finding antiderivatives in this way in Unit 7.
As shown in the example, a convenient way to carry out these sorts of
manipulations is to manipulate the expression inside the notation
/b
a . . . dx. This expression
/1 2
is known as the integrand. For example, the
definite integral 0 x dx has integrand x2 .
136
Example 5 Evaluating a more complicated definite integral

! 4
(2x + 1)(x − 4)
dx.
2 x
(It is shown in Figure 28.)
Solution
Manipulate the integrand to get it into a form that you can
integrate.
! 4
(2x + 1)(x − 4)
dx
2 x
! 4 2
2x − 7x − 4
= dx
2 x
! 4, -
4
= 2x − 7 − dx
2 x
! 4, -
1
= 2x − 7 − 4 × dx
2 x
Apply the fundamental theorem of calculus, and evaluate the
result.
) '4 Figure 28 The definite
= 2 × 12 x2 − 7x − 4 ln x 2
) '4 integral in Example 5
= x2 − 7x − 4 ln x 2
= (42 − 7 × 4 − 4 ln 4) − (22 − 7 × 2 − 4 ln 2)
= 16 − 28 − 4 ln 4 − 4 + 14 + 4 ln 2
= −2 − 4(ln 4 − ln 2)
= −2 − 4 ln(4/2)
= −2 − 4 ln 2
There’s no need to find a decimal approximation, as the question
didn’t ask for one.
There are other acceptable ways to leave the final answer in Example 5.
For example, you could write it as −2(1 + 2 ln 2).
137
Activity 10 Evaluating more complicated definite integrals
Evaluate the following definite integrals (which are shown in Figure 29).
Give exact answers.
! 4 ! 0 ! 4
√ r+5
(a) 3 x dx (b) x(1 + x) dx (c) dr
1 −1 2 r

When you use the square bracket notation, you can sometimes simplify
your working by using the following rules.
Constant multiple rule and sum rule for the square bracket
notation
) 'b ) 'b
kF (x) = k F (x) a , where k is a constant
a
) 'b ) 'b ) 'b
F (x) + G(x) a = F (x) a + G(x) a
For example, the constant multiple rule for the square bracket notation
tells you that
) '2 ) '2
5 sin x 1 = 5 sin x 1
and the sum rule for the square bracket notation tells you that
) '2 ) '2 ) '2
sin x + cos x 1 = sin x 1 + cos x 1 .
To see why these rules hold, you just need to use the definition of the
square bracket notation. For example,
) '2
5 sin x 1 = 5 sin 2 − 5 sin 1
= 5(sin 2 − sin 1)
) '2
= 5 sin x 1 .
138
Similarly,
) '2
sin x + cos x 1 = (sin 2 + cos 2) − (sin 1 + cos 1)
= (sin 2 − sin 1) + (cos 2 − cos 1)
) '2 ) '2
= sin x 1 + cos x 1 .
As always when a constant multiple rule and a sum rule hold, the sum rule
for the square bracket notation also holds if you replace the plus sign on
each side with a minus sign. To prove this, you combine the sum rule with
the case k = −1 of the constant multiple rule. You saw this done for the
constant multiple and sum rules for derivatives in Unit 6, Subsection 2.3.
The next example illustrates how you can use the constant multiple rule
and the sum rule for the square bracket notation when you’re evaluating a
definite integral.
Example 6 Using the constant multiple rule and sum rule for the
square bracket notation
! 2
(1 + x3 ) dx.
1
Solution
Use the fundamental theorem of calculus.
! 2
) '2
(1 + x3 ) dx = x + 14 x4 1
1
Use the sum rule.

) '2 ) '2
= x 1 + 14 x4 1
Figure 30 The definite
integral in Example 6
Use the constant multiple rule, and simplify the result.
) '2 ) '2
= x 1 + 14 x4 1
= (2 − 1) + 14 (24 − 14 )
1
=1+ 4 × 15
19
= 4 .
You can practise using the constant multiple rule and sum rule for the
square bracket notation in the next activity.
139
Activity 11 Using the constant multiple rule and sum rule for the
square bracket notation
Evaluate the following definite integrals. (They’re shown in Figure 31.)

Give your answer to part (a) to three significant figures.
! 5 ! 1
(a) x3/2 dx (b) (u − 2u2 ) du
3 −1/2

Note that when you’re evaluating a definite integral it’s sometimes more
convenient not to use the constant multiple rule and/or the sum rule for
the square bracket notation, even if there’s an opportunity to use them.
You can choose whether or not to use them, as seems convenient.
A constant multiple rule and a sum rule also hold for definite integrals, as
follows.
Constant multiple rule and sum rule for definite integrals

! b ! b
k f (x) dx = k
f (x) dx, where k is a constant
a a
! b ! b ! b
(f (x) + g(x)) dx = f (x) dx + g(x) dx
a a a
These rules follow from the fundamental theorem of calculus and the
constant multiple rule and sum rule for antiderivatives, which you met
in Unit 7. To see this for the constant multiple rule, suppose that f is a
continuous function whose domain contains the numbers a and b, and
that k is a constant. Let F be an antiderivative of f . Then, by the constant
multiple rule for antiderivatives, kF (x) is an antiderivative of kf (x).
140
Hence, by the fundamental theorem of calculus and the constant multiple

rule for the square bracket notation,
! b
) 'b
k f (x) dx = k F (x) a
a
) 'b
= k F (x) a
! b
=k f (x) dx.
a
You can prove the sum rule for definite integrals in a similar way.
Remember that the sum rule for definite integrals also holds if you replace
the plus sign on each side with a minus sign, because, as mentioned earlier,
this is always the case when a constant multiple rule and a sum rule hold.
The constant multiple rule for definite integrals tells you that, for example,
the area in Figure 32(b) is twice the area in Figure 32(a).
Figure 32 The area in (b) is twice the area in (a)

Similarly, the sum rule for definite integrals tells you that, for example, the
area in Figure 33(c) is the sum of the areas in Figure 33(a) and (b).
Figure 33 The area in (c) is the sum of the areas in (a) and (b)
The next example illustrates how you can use the constant multiple rule
and the sum rule for definite integrals when you’re evaluating a definite
integral.
141
Example 7 Using the constant multiple rule and sum rule for
definite integrals
! π
(x + 2 sin x) dx.
0
Solution
Use the fundamental theorem of calculus. Use the constant
multiple rule and sum rule, for definite integrals and for the square
bracket notation, as convenient.
! π ! π ! π
(x + 2 sin x) dx = x dx + 2 sin x dx
0 0 0
! π ! π
= x dx + 2 sin x dx
Figure 34 The definite 2 0 1π 0
) 'π
integral in Example 7 = 12 x2 + 2 − cos x 0
0
1
) 2 'π ) 'π
= 2 x 0 − 2 cos x 0
= 12 (π2 − 02 ) − 2(cos π − cos 0)
= 12 π2 − 2(−1 − 1)
= 12 π2 + 4.
You can practise using the constant multiple rule and sum rule for definite
integrals in the next activity.
Activity 12 Using the constant multiple rule and sum rule for
definite integrals
Evaluate the following definite integrals. (They are shown in Figure 35.)
Give exact answers.
! √7 ! π/4
1 3
(a) √ 2 x dx (b) (sin x + cos x) dx
3 −π/4
142

As with the sum rule and constant multiple rule for the square bracket
notation, you can choose whether or not to use the sum rule and the
constant multiple rule for definite integrals in your working, as seems
convenient. Often it’s more convenient not to use them, even if there’s an
opportunity to do so.
In the next activity, you’re asked to use the fundamental theorem of
calculus to find the exact area of the cross-section of a roof that was
discussed at the beginning of Subsection 1.1.
Activity 13 Using the fundamental theorem for a practical problem
The curve of the roof of a building is given by the graph of the function
1 2
f (x) = 9 − 36 x ,
where x and f (x) are measured in metres. Use the ideas that you’ve met
in this section to find the area between this graph and the x-axis from
x = −18 to x = 18, as shown below. Hence state the area of the
cross-section of the roof.
You may remember that in Subsection 1.1 we used subintervals to work

out that the area in Activity 13 is approximately 216 m2 . The solution to
Activity 13 confirms that the area is exactly 216 m2 .
Here’s another practical example for you to try. In this example the graph
lies below the x-axis, so the signed area that you’ll obtain by using the
fundamental theorem of calculus will be negative. You have to remove the
minus sign to obtain the required area.
143
Activity 14 Solving another practical problem about area
An engineer is planning the construction of an open channel with a

parabolic cross-section, as illustrated below. The shape of the cross-section
is given by the equation
1 2
y= 32 x − 200 (−100 ≤ x ≤ 100),
where x and y are measured in centimetres, as shown in the graph below.
The maximum height of liquid permissible in the channel is the height of

the x-axis on the graph; that is, 200 cm above the lowest point of the
channel. To ensure that the channel is adequate for the expected flow of
liquid, the engineer needs to know the area shaded on the graph.
(a) Find the x-intercepts of the graph.
(b) Hence calculate the shaded area, to the nearest square centimetre.
144
The history of the fundamental theorem of calculus

You’ve seen that the fundamental theorem of calculus relates rates of
change – that is, gradients of tangents – to signed areas.
One of the earliest mathematicians to relate the problem of
calculating tangents to the problem of calculating areas was
Isaac Barrow, the first Lucasian professor at the University of
Cambridge. However, his method was presented in the geometrical
style of the day and was not computationally useful for solving
problems. Isaac Newton was one of Barrow’s students, and in 1669
Barrow resigned his professorship so that Newton could take it up.
Newton based his definition of quadrature (determining area) on the
idea that it is the converse of finding tangents. So what for some Isaac Barrow (1630–1677)
people was, and for modern-day mathematicians is, a theorem that
finding areas and tangents are inverse processes – the fundamental
theorem of calculus – was taken by Newton as the very definition of
quadrature. Newton’s way of regarding area questions as the opposite
of tangency questions is set down clearly in his De analysi of 1669.
Gottfried Wilhelm Leibniz published his first accounts of differential
and integral calculus in the newly founded journal Acta Eruditorum
Lipsiensium (Acts of the scholars of Leipzig) in 1684 and 1686
respectively, and his proof of the fundamental theorem of calculus
followed in 1693.
The first rigorous proof of the fundamental theorem of calculus was
provided in the early decades of the 19th century by the French
mathematician Augustin Louis Cauchy, who included it in his course
at the École Polytechnique in Paris. Cauchy was one of the most
productive and influential mathematicians of his generation. His
reformulation of the foundations of calculus, while very thorough, was
also very hard to understand – so much so that not only his students
but also his fellow professors protested! Augustin Louis Cauchy
(1789–1857)
145
2.3 Using the fundamental theorem to find

changes in the values of antiderivatives
The fundamental theorem of calculus isn’t useful just for solving problems
related to finding areas. It’s also useful for solving problems involving rates
of change.
As you saw in Subsection 5.3 of Unit 7, in integral calculus you often want
to solve problems of the following type. You have a continuous function f ,
and two numbers a and b in its domain, and you want to work out the
amount by which an antiderivative of f changes from x = a to x = b.
(Remember that all the antiderivatives of f change by the same amount.)
If you can find a formula for an antiderivative F of f , then it’s
straightforward to do this: you just calculate F (b) − F (a). You saw
examples like this in Unit 7.
If you can’t find a formula for an antiderivative, then you can use the
fundamental theorem of calculus to help you find an approximate answer.
The theorem tells you that, for any antiderivative F of f ,
! b
F (b) − F (a) = f (x) dx.
a
So the answer that you want is equal to the signed area between the graph
of f and the x-axis, from x = a to x = b. You can use the method that you
met in Subsection 1.1 to find an approximate value for this signed area.
You would normally do that by using a computer, as mentioned earlier.
For example, suppose that you have a velocity-time graph for an object
moving along a straight line, such as the graph in Example 8 below. Since
displacement is an antiderivative of velocity, the fundamental theorem of
calculus tells you that the change in the displacement of the object, from
any point in time to any other point in time, is equal to the signed area
between the graph and the time axis, from the first point in time to the
second point in time. This fact is used in Example 8.
As with any graph, the units for signed area on a velocity-time graph are
the units on the vertical axis times the units on the horizontal axis. So, for
example, the units for signed area on the graph in Example 8 are
m s−1 × s = m;
that is, metres. These are units for displacement, as you’d expect.
Example 8 Using the fundamental theorem of calculus for a

velocity-time graph
The graph below is the velocity–time graph of an object moving along
a straight line. The areas (in m) of some regions are marked on the
graph.
146
In each of parts (a) to (c), use the given areas to find the amount by
which the displacement of the object changes from the first time to
the second time.
(a) From t = 0 to t = 4. (b) From t = 4 to t = 10.
(c) From t = 0 to t = 10.
Solution
(a) The change in displacement from t = 0 to t = 4 is 9 m.
(b) The change in displacement from t = 4 to t = 10 is −7 m.
(c) The change in displacement from t = 0 to t = 10 is
9 m + (−7) m = 2 m.
Notice that the velocity of the object in Example 8 changes from positive
to negative at time t = 4. That is, at this time the object changes from
moving in the positive direction to moving in the negative direction, as
illustrated in Figure 36. The negative displacement that occurs after this
time cancels out most of the positive displacement that occurred initially.
Since the object was displaced by 9 m in the positive direction, and then
by 7 m in the negative direction, its final displacement from its starting
position is 2 m, as found in Example 8(c). The object travels a total
distance of 9 m + 7 m = 16 m during the 10 seconds of its motion.
Figure 36 The object in Example 8 starts moving in the positive

direction, then turns and moves in the negative direction
147
Activity 15 Using the fundamental theorem of calculus for a

velocity–time graph
The velocity–time graph of an object moving along a straight line is shown

below. The areas of some regions are marked on the graph. Use these
areas to find the answers in parts (a) to (c) below.
(a) In each of parts (i) to (iv), find the amount by which the displacement
of the object changes from the first time to the second time.
(i) From t = 0 to t = 1. (ii) From t = 3 to t = 4.
(iii) From t = 0 to t = 3. (iv) From t = 3 to t = 6.
(b) What is the displacement of the object from its starting position at
each of the following times?
(i) t = 2 (ii) t = 6
(c) What is the total distance travelled by the object during the 6 seconds
of its motion?
If you want to solve a problem that involves finding a change in the value
of an antiderivative, and you can find a formula for an antiderivative, then
it’s usually convenient to use the notation for a definite integral, and the
square bracket notation, in your working. Here’s Example 22 from
Subsection 5.3 of Unit 7 again, but with the working carried out using this
notation.
148
Example 9 Finding a change in the value of an antiderivative

Suppose that the rate of change of a quantity is given by the function
f (x) = 15 x2 (x ≥ 0), as shown in Figure 37. By what amount does the
quantity change from x = 3 to x = 6?
Solution
The change in the quantity from x = 3 to x = 6 is
! 6 2 16
1 2 1 1 3
5 x dx = 5 × 3 x
3 3
2 16
1 3
= 15 x
3
1
) '
3 6
= 15 x 3
1 3 3
= 15 (6 − 3 )
1
= 15 × 189 Figure 37 The graph of
63
= 5 y = 15 x2 (x ≥ 0)
= 12.6.
In the next activity, you’re asked to repeat Activity 36 from Subsection 5.3
of Unit 7, but using the notation for a definite integral.
Activity 16 Finding a change in displacement
Suppose that an object moves along a straight line, and its velocity v
(in m s−1 ) at time t (in seconds) is given by v = 3 − 32 t (0 ≤ t ≤ 6)
(as shown in Figure 38).
(a) Write down a definite integral that gives the change in the
displacement of the object from time t = 0 to time t = 1. Evaluate it,
and hence state how far the object travels in that time.
(b) Repeat part (a), but from time t = 4 to time t = 5.
Try using the notation for a definite integral to solve the problem in the
next activity.
Figure 38 The graph of

v = 3 − 32 t (0 ≤ t ≤ 6)
149
Activity 17 Finding another change in displacement
The driver of a car travelling at 28 m s−1 along a straight road applies the
brakes. It takes four seconds for the car to stop, and its decreasing
velocity v (in m s−1 ) during that time is given by the equation
v = 28 − 7t,
where t is the time in seconds since the brakes were applied (as shown in
Figure 39). How far does the car travel during those four seconds?
In this subsection you’ve seen examples of the following two useful facts.
For the velocity-time graph of any object, and for any two points in
Figure 39 The graph of time t = a and t = b in the time period covered by the graph, the
v = 28 − 7t (0 ≤ t ≤ 4) following hold.
• The change in the displacement of the object from t = a to t = b
is the total signed area between the graph and the time axis from
t = a to t = b.
• The total distance travelled by the object from t = a to t = b is
the total area between the graph and the time axis from t = a to
t = b (this applies when a ≤ b).
Note that if a graph is a straight line, or consists of several line segments,

then integration may not be the easiest way to work out an area or a
signed area between the graph and the horizontal axis. For example, you
can work out the area shaded in Figure 39 by using the formula for the
area of a triangle, as shown at the end of the solution to Activity 17. Try
using a similar geometric method in the next activity, which is a repeat of
Example 23 in Unit 7.
Activity 18 Finding yet another change in displacement
Suppose that a car begins to move along a straight road, and its velocity v
(in m s−1 ) during the first five seconds of its journey is given by the
equation
v = 3t,
where t is the time in seconds since it began moving, as shown in
Figure 40. How far does the car travel from the end of the fourth second to
the end of the fifth second?
v = 3t (0 ≤ t ≤ 5)
150
2.4 Notation for indefinite integrals

Until now, when we’ve been working with antiderivatives and indefinite
integrals, we’ve been using a lower-case letter, such as f , to denote the
function that we start with, and the corresponding upper-case letter, such
as F , to denote any of its antiderivatives, or its indefinite integral.
For example, some antiderivatives of the function f (x) = x2 are
F (x) = 31 x3 , F (x) = 13 x3 + 7, and F (x) = 13 x3 − 29 ,
and the indefinite integral of this function f is
F (x) = 13 x3 + c.
However, there’s a standard notation for indefinite integrals, which arises

from the link between antiderivatives and definite integrals. For any
function f that has an antiderivative, the indefinite integral of f (x) is
denoted by
!
f (x) dx.
(This notation is read as ‘the integral of f of x, d x’.)

For example,
!
x2 dx = 13 x3 + c,
and similarly
!
sin x dx = − cos x + c.
(You saw in Unit 7 that the indefinite integral of sin x is − cos x + c.)
Note that this notation is used only for indefinite integrals – it’s never used
for antiderivatives. So it’s important to remember the following fact.
/
If an equation has the notation . . . dx on one side only, then there
must be an arbitrary constant on the other side.
For example, the equation

!
x2 dx = 13 x3
is incorrect. The correct equation is

!
x2 dx = 13 x3 + c,
as you saw above.

We’ll continue to use the capital letter notation for antiderivatives, when
it’s convenient to do so.
151
The notation for indefinite integrals introduced above, and the similar
notation for definite integrals, are together known as integral notation.
We refer to either an indefinite integral or a definite integral as an
integral. In the indefinite integral
!
f (x) dx,
the expression f (x) is called the integrand, just as in a definite integral.

The next example illustrates how integral notation is used for indefinite
integrals in practice. It’s similar to the examples and activities on finding
indefinite integrals in the second half of Unit 7, except that it uses integral
notation.
Example 10 Finding indefinite integrals, using integral notation

Find the following indefinite integrals.
! !
2 x−5
(a) sec x dx (b) dx
x
Solution
(a) The indefinite integral of sec2 x is given in the table of
standard indefinite integrals, so you can simply write it down.
!
sec2 x dx = tan x + c
(b) Manipulate the integrand to get it into a form that allows you
to use the constant multiple rule and sum rule for antiderivatives,
together with standard indefinite integrals.
! ! , -
x−5 5
dx = 1− dx
x x
! , -
1
= 1−5× dx
x
= x − 5 ln |x| + c
152
You can practise using integral notation for indefinite integrals in the
following activity. Don’t forget to add the ‘ + c’ in each part.
Activity 19 Finding indefinite integrals, using integral notation

! ! !
1
(a) ex dx (b) (x + 3)(x − 2) dx (c) dx
x3
! ! !
1 −3
(d) (sin θ − cos θ) dθ (e) 2
dx (f) du
1+x 1 + u2
! !
1 (1 + x2 )
(g) dr (h) dx
2(1 + r2 ) 2
Since an indefinite integral is just an antiderivative with an arbitrary

constant added on, the constant multiple rule and sum rule for
antiderivatives, from Unit 7, can be written in terms of indefinite integrals.
These forms of the rules are stated in the following box, in integral
notation.
Constant multiple rule and sum rule for indefinite integrals

! !
k f (x) dx = k f (x) dx, where k is a constant
! ! !
(f (x) + g(x)) dx = f (x) dx + g(x) dx
As with the other constant multiple rules and sum rules, you can choose
whether or not to use these rules in your working, as seems convenient.
The next example repeats Example 10(b), but uses the constant multiple
rule and sum rule for indefinite integrals in the forms stated above.
153
Example 11 Using the constant multiple rule and sum rule for
indefinite integrals
Find the indefinite integral
!
x−5
dx.
x
Solution
Manipulate the integrand to get it into a form that allows you to
use the constant multiple rule and sum rule for indefinite integrals,
together with standard indefinite integrals.
! ! , -
x−5 5
dx = 1− dx
x x
! !
1
= 1 dx − 5 dx
x
= x − 5 ln |x| + c
Here’s something to notice about Example 11. You might have thought
that each of the two indefinite integrals in the second line of the working
should have been replaced by an expression containing an arbitrary
constant. That would lead to a solution something like this:
! ! , -
x−5 5
dx = 1− dx
x x
! !
1
= 1 dx − 5 dx
x
= (x + c) − 5(ln |x| + d)
= x − 5 ln |x| + c − 5d,
where c and d are arbitrary constants.
However, since the arbitrary constants c and d can take any value, the
expression c − 5d can also take any value, so we may as well replace it by a
single letter that can take any value. It’s convenient to use the usual
letter, c. In general, if an expression consists of a sum of constant multiples
of indefinite integrals, then it needs only a single arbitrary constant.
You can practise using the constant multiple rule and sum rule for
indefinite integrals in the next activity.
154
3 Integration by substitution
Activity 20 Using the constant multiple rule and sum rule for
indefinite integrals

! , - !
3 1
(a) − dx (b) (cosec x)(cosec x + cot x) dx
x 1 + x2
You’ve now reached the end of the material in this module that tells you
what integration is all about. In the rest of this unit you’ll concentrate on
actually finding indefinite and definite integrals.
Before you go on to that, you might like to watch the 25-minute video
The birth of calculus, which is available on the module website. It tells you
more about the history of the development of calculus.
To allow you to fully exploit all the ideas about integration that you’ve
met, it’s important that you’re able to find formulas for the indefinite
integrals of as wide a range of functions as possible. In the rest of this
unit, you’ll learn to integrate a much wider variety of functions than you
learned to integrate in Unit 7. In the final subsection of the unit, you’ll
learn how to use the module computer algebra system to find indefinite
and definite integrals.
Unfortunately, as mentioned in Unit 7, it’s generally more tricky to
integrate functions than it is to differentiate them. That’s because there
are no rules for antiderivatives that are similar to the product, quotient
and chain rules for derivatives, so you can’t usually integrate functions in
the systematic way that you can differentiate them.
However, there are still many useful techniques that you can use to
integrate functions. Each of these techniques applies to functions with
particular characteristics. You’ll meet some of these techniques in the rest
of this unit. You’ll learn how to recognise functions to which each
technique can be applied, and how to use the techniques.
The technique that you’ll learn in this section is integration by substitution.
This technique might seem quite complicated when you first meet it, but
you should find it more straightforward once you’ve had some practice
with it. If you find it difficult, then make sure that you watch the tutorial
clips, and try some extra practice. Remember that there are more practice
exercises in the practice quiz and exercise booklet for this unit.
155
3.1 Basic integration by substitution

Integration by substitution is based on reversing the chain rule for
differentiation.
Consider what happens when you differentiate an expression by using the
chain rule. Remember that you start by recognising that the expression is
a function of ‘something’. You differentiate the expression with respect to
the ‘something’, then you multiply the result by the derivative of the
‘something’ with respect to the input variable (usually x). For example, by
the chain rule,
d & + & +
sin(x2 ) = cos(x2 ) (2x).
dx
Since differentiation is the reverse of integration, this equation tells you
that
!
& +
cos(x2 ) (2x) dx = sin(x2 ) + c.
Whenever you differentiate an expression by using the chain rule, you

always obtain an expression of the form
f (something) × the derivative of the something,
where f is a function that you can integrate. The essential idea of
integration by substitution is to recognise an expression that you want to
integrate as having this form, and then perform the integration by
reversing the chain rule.
To help you reverse the chain rule, it’s helpful to denote the ‘something’ by
an extra variable, in the way that you did when you first learned to use the
chain rule in Unit 7. The usual choice of letter for the extra variable is u.
Here’s how you can use this method to find the integral
!
& +
cos(x2 ) (2x) dx,
if you hadn’t first seen the integrand obtained as an ‘output’ of the chain
rule.
The first step is to recognise that the integrand has the form
where f is a function that you can integrate. You can see that it does,
since it is
cos(something) × the derivative of the something,
where the ‘something’ is x2 .
Then, since the ‘something’ is x2 , you put u = x2 . This gives du/dx = 2x.
Hence, in the integral you can replace cos(x2 ) by cos u, and 2x by du/dx,
to give
!
du
cos(u) dx.
dx
156
Now here’s the crucial step that you need to apply at this stage. Any
integral of the form
!
du
f (u) dx
dx
is equal to the simpler integral
!
f (u) du.
This is the step that uses the chain rule, and you’ll see an explanation of
why it’s correct later in this subsection. Notice that it’s easy to remember,
because it looks like we’ve simply cancelled a ‘dx’ in a denominator with a
‘dx’ in a numerator. (Of course that’s not what we’ve done, since du/dx
isn’t a fraction.)
So the integral that we’re trying to find here can be expressed in terms
of u as the simpler integral
!
cos u du,
where u = x2 . You can now do the integration, which gives

sin u + c.
The final step is to express this answer in terms of x, using the fact that
u = x2 . This gives the final answer
sin(x2 ) + c.
So we’ve now worked out that
!
& +
cos(x2 ) (2x) dx = sin(x2 ) + c,
as expected.
Here’s a summary of the method used above, which is the method known
as integration by substitution.
Integration by substitution
1. Recognise that the integrand is of the form
where f is a function that you can integrate.
2. Set the something equal to u, and find du/dx.
3. Hence write the integral in the form
!
f (u) du,
! !
du
by using the fact that f (u) dx = f (u) du.
dx
4. Do the integration.
5. Substitute back for u in terms of x.
157
As illustrated by the example above, the effect of substituting the

variable u in place of the ‘something’ in the integral is to reduce the
integral to a simpler integral in terms of u, which you may be able to find
more easily.
You’ll see another example of integration by substitution shortly, and
you’ll then have a chance to try it for yourself, but first it’s useful for you
to learn about a convenient way to carry out step 3 of the method.
Consider once more the integral above:
!
& +
cos(x2 ) (2x) dx.
Remember that we put u = x2 , which gave du/dx = 2x, and this allowed
us to write the integral in terms of u as
!
cos u du.
It’s convenient to think of this new form of the integral as being obtained
from the original form by making two replacements, as shown below:
!
& +
cos(x2 ) (2x) dx .
* %$ 3 * %$ 3
cos u du
It’s straightforward to work out that you can replace cos(x2 ) by cos u, as
that just comes from the equation u = x2 .
A helpful way to work out that you can replace (2x) dx by du is to imagine
‘cross-multiplying’ in the equation
du
= 2x,
dx
to obtain
du = (2x) dx.
This isn’t a real equation, of course, since du and dx don’t have
independent meanings outside the notation for a derivative or integral.
But it tells you immediately that you can replace (2x) dx by du. (In fact,
it’s possible to attach a meaning to du and dx, and hence to the equation
above, but this is outside the scope of this module.)
Here’s another example of integration by substitution, which illustrates all
the ideas above.
Example 12 Integrating by substitution

Find the integral
!
etan x sec2 x dx.
158
Solution
Since the derivative of tan x is sec2 x, the integrand is of the form
esomething × the derivative of the something.
So set the something equal to u.
du
Let u = tan x; then = sec2 x.
dx
Imagine ‘cross-multiplying’ in the second equation to obtain
du = (sec2 x) dx.
So
! !
tan x 2
e sec x dx = eu du
Do the integration.
= eu + c
Substitute back for u in terms of x.
= etan x + c.
Here are some examples for you to try. Make sure that you have your
Handbook to hand, so you can consult the tables of standard derivatives
and standard integrals.
Activity 21 Integrating by substitution

! !
& +
(a) ecos x (− sin x) dx (b) sin(x3 ) (3x2 ) dx
! , - ! , -
1 1
(c) cos x dx (d) (− sin x) dx
sin x cos2 x
! ! , -
4 1
(e) sin x cos x dx (f) cos x dx
1 + sin2 x
In the examples that you’ve seen so far, it’s been fairly straightforward to
decide what the ‘something’ should be, so that the integrand is of the form
f (something) × the derivative of the something.
However, sometimes you need to think about this more carefully, as
illustrated in the next example.
159
Example 13 Choosing the ‘something’ when you integrate by

substitution
Find the integral
! , -
1
cos x dx.
2 + sin x
Solution
The integrand is of the form
1
× the derivative of the something,
2 + something
where the something is sin x.
It’s also of the form
1
× the derivative of the something,
something
where the something is 2 + sin x.
Choosing the first option would lead to the first part of the integrand
being replaced by 1/(2 + u), which isn’t straightforward to integrate.
Choosing the second option would lead to the first part of the
integrand being replaced by 1/u, which you can integrate using a
result from the table of standard integrals.
So choose the second option: take the something to be 2 + sin x, and
set it equal to u.
du
Let u = 2 + sin x; then = cos x.
dx
Imagine ‘cross-multiplying’ to obtain du = cos x dx.
So
! , - !
1 1
cos x dx = du
2 + sin x u
= ln |u| + c
= ln |2 + sin x| + c
In the particular case here you can remove the modulus signs,
since 2 + sin x is always positive.
= ln(2 + sin x) + c.
In each part of the next activity it’s useful to take the ‘something’ to be an
expression that includes a constant term, in a similar way to Example 13.
160
Activity 22 Choosing the ‘something’ when you integrate by

substitution
! ! "
7
(a) (4 + cos x) (− sin x) dx (b) 1 + x2 (2x) dx
! ! , -
5 10 4 1
(c) (x − 8) (5x ) dx (d) ex dx
ex + 5
!
& +
(e) sin(5 + 2x3 ) (6x2 ) dx
In the next activity, some of the integrals are similar to those in

Activity 22, whereas others are similar to those in Activity 21. You have
to decide in each case what the ‘something’ should be. Remember that the
idea is that when you replace the something by u in the first part of the
integrand, you obtain an expression in u that you can integrate.
Activity 23 Choosing the ‘something’ when you integrate by

substitution
! !
x x
(a) (cos(e )) e dx (b) e1+sin x (cos x) dx
! , - !
1 9
(c) (10x ) dx (d) (cos3 x)(− sin x) dx
x10 + 6
Finally in this subsection, as promised near the start, here’s an

explanation of why
! !
du
f (u) dx = f (u) du,
dx
where the variable u is a function of x, and f is a function that you can
integrate. To justify this, we let F be an antiderivative of f , and apply the
chain rule to F (u). This gives
d & + du
F (u) = f (u) .
dx dx
It follows from this equation that
!
du
f (u) dx = F (u) + c.
dx
However, because F is an antiderivative of f , we also have
!
f (u) du = F (u) + c.
Comparing the final two equations above proves the result.

161
3.2 Integration by substitution in practice

Usually when you have an integrand that you can integrate by
substitution, it won’t initially be exactly in the form
You have to start by recognising that it can be rearranged into this form,
then rearrange it accordingly. Here are two examples that illustrate some
things that you can do.
Example 14 Rearranging an integral before using substitution

Find the integral
!
ex cos(ex ) dx.
Solution
The derivative of ex is ex , so the integrand is in the required form,
except that the two factors cos(ex ) and ex are in the ‘wrong’ order. If
you swap them, then the integrand will be exactly in the required
form.
du
Let u = ex ; then = ex .
dx
Imagine ‘cross-multiplying’ to obtain du = ex dx. Then start by
swapping the two factors in the integrand.
So
! !
x x
e cos(e ) dx = (cos(ex )) ex dx
!
= cos u du
= sin(u) + c
= sin(ex ) + c.
162
Example 15 Rearranging another integral before using substitution

Find the integral
!
& +8
x2 1 + x3 dx.
Solution
The derivative of 1 + x3 is 3x2 , which is ‘nearly’ x2 (it just has the
wrong coefficient). So if you swap the order of the two factors x2 and
(1 + x3 )8 , then the integrand will be nearly in the required form.
du
Let u = 1 + x3 ; then = 3x2 .
dx
Imagine ‘cross-multiplying’ to obtain du = 3x2 dx. Then start by
swapping the two factors in the integrand.
So
! !
& +8 & +8
x 1 + x3 dx =
2
1 + x3 x2 dx
The only problem is that in place of x2 you need 3x2 . To achieve

this, multiply by 3 inside the integral, and divide by 3 outside, to
compensate. (The constant multiple rule for indefinite integrals tells
you that it’s okay to do this.)
!
& +8
=3 1
1 + x3 (3x2 ) dx
Now do the substitution.

!
= 3 u8 du
1
1 1 9
= 3 × 9u + c
1 9
= 27 u + c
1 3 9
= 27 (1 + x ) + c.
Here’s something to notice

/ about Example 15. You might have thought
that when the integral u8 du in the fourth-last line of the solution was
replaced by 91 u9 + c, the solution should have proceeded like this:
!
= 3 u8 du
1
& +
= 13 19 u9 + c
1 9 1
= 27 u + 3 c
1 3 9
= 27 (1 + x ) + 13 c.
163
However, since the arbitrary constant c can take any value, the expression
1
3 c can also take any value, so we may as well replace it by a single letter
that can take any value. It’s convenient to use the usual letter, c.
This sort of thing occurs frequently when you integrate. You never need to
multiply or divide a term that is an arbitrary constant by a number – you
just replace it by another arbitrary constant, and you usually use the
standard letter, c.
Another point to remember from the working in Example 15 is the
strategy of multiplying by a constant inside an integral, and dividing by
the same constant outside to compensate. It’s often helpful to do this
when you use integration by substitution. It’s okay to do it because the
constant multiple rule for indefinite integrals tells you that for any
function f that has an antiderivative, and any non-zero constant k,
! ! !
1 1
f (x) dx = kf (x) dx = kf (x) dx.
k k
Remember that the constant k can be either positive or negative. In
particular, it can be −1.
Remember too that k must be a constant. You can’t do the same thing
with an expression in x.
Here are some examples for you to try.
Activity 24 Rearranging integrals before using substitution
Find the following integrals.

! ! " !
2 3
(a) x cos(x ) dx (b) x 1 + x2 dx (c) cos4 x sin x dx
The next example illustrates another way in which you can sometimes
rearrange an integrand to get it into the right form for integration by
substitution.
Example 16 Rearranging another integral before using substitution

Find the integral
!
x
dx.
1 − x2
164
Solution
The derivative of the denominator 1 − x2 is −2x, which is ‘nearly’
the numerator x (it just has the wrong coefficient). So if you split the
numerator from the rest of the fraction, then the integrand will be
nearly in the required form.
du
Let u = 1 − x2 ; then = −2x.
dx
Imagine ‘cross-multiplying’ to obtain du = −2x dx. Then start by
splitting the numerator from the rest of the fraction in the
integrand.
So
! ! , -
x 1
dx = x dx
1 − x2 1 − x2
The only problem is that in place of x you need −2x. To achieve
this, multiply by −2 inside the integral, and divide by −2 outside to
compensate.
! , -
1 1
= −2 (−2x) dx
1 − x2
Now do the substitution.
!
1 1
= −2 du
u
= − 12 ln |u| + c
= − 12 ln |1 − x2 | + c.
The type of integral in Example 16 occurs quite frequently, so it’s worth

learning to recognise it. You can use integration by substitution to find
any integral of the form
!
the derivative of the something
dx,
something
or
!
the derivative of the something
dx, where n is a constant.
(something)n
The constant n can be any nonzero real number. Here are some integrals
of this type for you to find.
165
Activity 25 Rearranging more integrals before using substitution

! ! !
x ex cos x
(a) dx (b) dx (c) dx
1 + 2x2 (2 + ex )2 3 − sin x
Here are three more integrals for you to find by using substitution.
They’re a mixture of the types in Activities 24 and 25.
Activity 26 Rearranging even more integrals before using

substitution
! ! !
ex
(a) ecos x sin x dx (b) dx (c) (1 + 12 sin x)2 cos x dx
3 − ex
You can use integration by substitution to find the indefinite integral of

the tangent function. You’re asked to do that in the next activity.
Activity 27 Finding the indefinite integral of tan
By writing
sin x
tan x = ,
cos x
and using integration by substitution, find
!
tan x dx.
Remember that when you’ve found an indefinite integral, you can always
check your answer by differentiating it. This can be particularly helpful
when you found the integral by using integration by substitution.
166
3.3 Integrating functions of linear expressions

This subsection is about a particular type of integral that occurs
frequently, and which you can find by using integration by substitution.
The integrals
! ! !
1
sin(5x − 2) dx, dx, and e−9x dx
4−x
all have integrands of the form f (ax + b), where f is a function, and
a and b are constants with a (= 0. In other words, each of the integrands is
of the form f (a linear expression in x).
You can always find an integral of this form by using integration by
substitution, provided that you can integrate the function f . Here’s an
example.
Example 17 Integrating a function of a linear expression

Find the integral
!
sin(5x − 2) dx.
Solution
The derivative of 5x − 2 is 5, so the integral will be in the required
form if you rearrange it to obtain 5 inside.
du
Let u = 5x − 2; then = 5. So
dx
! !
sin(5x − 2) dx = 15 sin(5x − 2) × 5 dx
!
= 15 sin(5x − 2) × 5 dx
!
1
= 5 sin u du
= − 15 cos u + c
= − 15 cos(5x − 2) + c.
167
Here are some similar integrals for you to find.
Activity 28 Integrating functions of linear expressions

! ! !
6x−1
(a) e dx (b) sin(4x) dx (c) e−9x dx
! ! !
−x/3 1
(d) e dx (e) cos(3 − 7x) dx (f) dx
4−x
! ! !
1 1
(g) dx (h) dx (i) sec2 (6x) dx
(2x + 1)2 (1 − x)3
The next example uses exactly the same ideas as Example 17 and
Activity 28, but is a little more complicated.
Example 18 Integrating another function of a linear expression

Find the integral
!
1
dx.
1 + 9x2
Solution
1
You can write the integrand as , which is of the
1 + (3x)2
1
form f (3x) where f (x) = . The integral of this function f is a
1 + x2
standard integral. Since the derivative of 3x is 3, you need to
rearrange the given integral to obtain 3 inside.
du
Let u = 3x; then = 3. So
dx
! !
1 1
2
dx = dx
1 + 9x 1 + (3x)2
!
1
= 13 × 3 dx
1 + (3x)2
!
1 1
=3 du
1 + u2
= 13 tan−1 u + c
= 1
3 tan−1 (3x) + c.
168
Here are some similar integrals for you to find.
Activity 29 Integrating more functions of linear expressions

! ! !
1 1 1
(a) √ dx (b) dx (c) dx
1 − 4x2 1 + 2x2 4 + 9x2
Hint for part (c): start by writing
1 1 1
as × .
4 + 9x2 4 1 + 94 x2
The answers to Activities 28 and 29 all follow the same general pattern, as
follows. Suppose that you have an integral of the form
!
f (ax + b) dx,
where a and b are constants with a (= 0, and you know an antiderivative F

of the function f . To find the integral by substitution, you put u = ax + b,
du
which gives = a. Then
dx
! !
1
f (ax + b) dx = f (ax + b) a dx
a
!
1
= f (u) du
a
1
= F (u) + c
a
1
= F (ax + b) + c.
a
So we have the following general fact.
Indefinite integral of a function of a linear expression

If f is a function with antiderivative F , and a and b are constants
with a (= 0, then
!
1
f (ax + b) dx = F (ax + b) + c.
a
It’s worth remembering this fact, and using it to deal with integrals where
the integrand is a simple function of a linear expression, rather than using
integration by substitution every time. For example, the fact above tells
you immediately that
!
cos(3x − 2) dx = 13 sin(3x − 2) + c.
169
The following special case of the fact in the box above is useful especially
often.
Indefinite integral of a function of a multiple of the variable

If f is a function with antiderivative F , and a is a non-zero constant,
then
!
1
f (ax) dx = F (ax) + c.
a
For example,
!
e−2x dx = − 12 e−2x + c.
Try using the facts in the boxes above in the following activity.
Activity 30 Integrating functions of linear expressions efficiently

! ! !
(a) cos(10x) dx (b) sin(2 − 5x) dx (c) e3t+4 dt
! ! !
1
(d) cos( 12 θ) dθ (e) e−4p dp (f) dx
7x − 4
! ! !
1 2
(g) dr (h) sec (5x − 1) dx (i) sin( 15 x) dx
1−r
! ! !
−x x/2
(j) e dx (k) e dx (l) sec2 (2 − 3p) dp
! , -
2−x
(m) sin dx
3
The next example is of a type that you met in Unit 7, but it involves a
function that you’ve learned how to integrate in this section.
170
Example 19 Integrating in a practical situation

The velocity of an object moving along a straight line is given by the
equation
& +
v = 12 sin 32 t − 2 ,
where t is the time in seconds since the object began moving, and v is
the velocity of the object, in m s−1 . The displacement of the object at
time t = 0 is 3 m.
Find an equation for the displacement s (in metres) of the object, in
terms of t. Give the constant term to two decimal places.
Solution
Use the fact that displacement is an antiderivative of velocity.
The velocity v (in m s−1 ) of the object at time t (in seconds) is given
by
& +
v = 12 sin 32 t − 2 .
Hence the displacement s (in m) of the object at time t (in seconds) is
given by
!
1
&3 +
s= 2 sin 2 t − 2 dt
1 1 & & ++
= × − cos 32 t − 2 + c
2 3/2
& +
= − 13 cos 32 t − 2 + c,
where c is an arbitrary constant.
Use information given in the question to find the value of c.
When t = 0, s = 3. Hence
& +
3 = − 13 cos 32 × 0 − 2 + c
3 = − 13 cos(−2) + c
c = 3 + 13 cos(−2)
c = 3 + 13 cos 2
c = 2.86 (to 2 d.p.).
So the equation for the displacement of the object in terms of time is
& +
s = − 13 cos 32 t − 2 + 2.86.
171
Activity 31 Integrating in a practical example
The velocity of an object moving along a straight line is given by the

equation
& +
v = 2 cos 12 t + 5 ,
where t is the time in seconds since the object began moving, and v is the
velocity of the object, in m s−1 . The displacement of the object at time
t = 0 is 4 m.
(a) Find an equation for the displacement s (in metres) of the object, in
terms of t. Give the constant term to two decimal places.
(b) Find the displacement of the object at time t = 10. Give your answer
in metres to two significant figures.
3.4 Integration by substitution for definite integrals

Once you’ve used integration by substitution to find an indefinite integral,
you can find any definite integral with the same integrand by using the
fundamental theorem of calculus in the usual way. Here’s an example.
Example 20 Evaluating a definite integral by substitution (first

method)
! 3π/2
esin x cos x dx.
π/2
(It’s shown in Figure 41.) Give your answer to three significant

figures.
Solution
Use integration by substitution to obtain the indefinite integral of
the integrand.
Figure 41 The definite du
integral in Example 20 Let u = sin x, then = cos x. So
dx
! !
esin x cos x dx = eu du
= eu + c
= esin x + c.
172
Apply the fundamental theorem of calculus.

Hence
! 3π/2
) '3π/2
esin x cos x dx = esin x π/2
π/2
= esin(3π/2) − esin(π/2)
= e−1 − e1
1
= −e
e
= −2.35 (to 3 s.f.).
An alternative way to use integration by substitution to evaluate a definite

integral is to work with the definite integral from the start, rather than
first finding an indefinite integral. If you do this, then because the limits of
integration are values of x, when you change the variable from x to u you
must also change the limits of integration to the corresponding values of u.
This method is demonstrated in the next example. The definite integral in
this example is the same as the one in Example 20.
Example 21 Evaluating a definite integral by substitution (second

method)
! 3π/2
esin x cos x dx.
π/2
Give your answer to three significant figures.

Solution
Use integration by substitution.
du
Let u = sin x, then = cos x.
dx
Find the values of u that correspond to the values of x that are
the limits of integration.
.π#
Putting x = π/2 into u = sin x gives u = sin = 1.
2
, -
3π
Putting x = 3π/2 into u = sin x gives u = sin = −1.
2
173
Change the variable from x to u, remembering that you also have

to change the limits of integration.
So
! 3π/2 ! −1
sin x
e cos x dx = eu du
π/2 1
) '−1
= eu 1
= e−1 − e1
= −2.35 (to 3 s.f.).
You can try using this method in the next activity.
Activity 32 Evaluating definite integrals by substitution
Evaluate the following definite integrals. Give your answers to three

! 1 ! π ! 0
x 3 2
(a) 2
dx (b) cos x sin x dx (c) xe−3x dx
0 1 + 2x π/2 −1
3.5 Finding more complicated integrals by

substitution
You can sometimes use integration by substitution to find an integral even
if the integrand isn’t obviously of the form
In this subsection, you’ll see some examples where the integrand is nearly
of this form, except that it also contains a further factor, which is an
expression in x. If you can express this further factor in terms of the
variable u that you use for the substitution, then you may be able to
simplify the integrand into a form that you can integrate. Here’s an
example.
174
Example 22 Expressing a further factor in terms of u when

integrating by substitution
Find the integral
!
(x + 1)(3x + 1)9 dx.
Solution
If the integral were just
!
(3x + 1)9 dx,
then we could use the substitution u = 3x + 1, and find the integral

by first writing it as
!
1
3 (3x + 1)9 × 3 dx.
However, the integrand contains a further factor, x + 1. Try using the

substitution, expressing the further factor x + 1 in terms of u.
du
Let u = 3x + 1. Then = 3.
dx
Also, rearranging the equation u = 3x + 1 gives
u = 3x + 1
3x = u − 1
x = 13 (u − 1).
Hence
x + 1 = 13 (u − 1) + 1
= 13 (u + 2).
So
! !
9
(x + 1)(3x + 1) dx = 1
3 (x + 1)(3x + 1)9 × 3 dx
!
= 1
3
1
3 (u + 2)u9 du
!
= 1
9 (u + 2)u9 du
!
= 1
9 (u10 + 2u9 ) du
&
1 1 11 1 10
+
= 9 11 u + 2 × 10 u +c
1 11 1 10
= 99 u + 45 u + c
1 11 1 10
= 99 (3x + 1) + 45 (3x + 1) + c.
175
Here are two similar integrals for you to try.
Activity 33 Expressing further factors in terms of u
Use integration by substitution to find the following indefinite integrals.

! !
√ x−2
(a) x 1 + x dx (b) dx
(x + 3)3
You’ll learn more about integration by substitution in the module MST125

Essential mathematics 2, if it’s in your study programme.
4 Integration by parts
In this section you’ll meet another integration technique, integration by
parts. Like integration by substitution, this technique is useful for
integrating functions that have certain characteristics. It’s based on the
product rule for differentiation.
4.1 Basic integration by parts

Suppose that f and g are functions, and that the function G is an
antiderivative of the function g. Consider the equation that you obtain
when you use the product rule to differentiate the product f (x)G(x):
, - , -
d d d
(f (x)G(x)) = f (x) G(x) + G(x) f (x) ,
dx dx dx
that is,
d
(f (x)G(x)) = f (x)g(x) + G(x)f ! (x).
dx
If you swap the order of G(x) and f ! (x) in the final term, and subtract this
term from both sides of the equation, then you obtain
d
(f (x)G(x)) − f ! (x)G(x) = f (x)g(x).
dx
Swapping the sides of this equation gives
d
f (x)g(x) = (f (x)G(x)) − f ! (x)G(x).
dx
176
If you now integrate both sides of this equation with respect to x, then you
obtain
! !
f (x)g(x) dx = f (x)G(x) − f ! (x)G(x) dx.
There’s no need to add an

/ explicit arbitrary constant here, because it’s
included/in the integral f (x)g(x) dx on the left-hand side, and in the
integral f ! (x)G(x) dx on the right-hand side.
The formula above is known as the integration by parts formula. It’s
stated again in the box below, for easy reference. If you’ve met it before,
then you may have seen it stated in a different form. This is explained at
the end of this section.
Integration by parts formula (Lagrange notation)

! !
Here G is an antiderivative of g.
The integration by parts formula is useful for integrating some products of

the form f (x)g(x). You can see that it changes the problem of integrating
f (x)g(x) into the problem of integrating f ! (x)G(x), where G is an
antiderivative of g.
So it’s useful for integrating products of the form f (x)g(x) that have the
following characteristics.
• You can find an antiderivative G of g; in other words, you can
integrate the expression g(x).
• The product f ! (x)G(x) is easier to integrate than the product
f (x)g(x). For example, this may be the case if f ! (x) is simpler than
f (x).
When we integrate by using the integration by parts formula, we say that
we’re integrating by parts. Here’s an example.
177
Example 23 Integrating by parts

!
Find the integral x sin x dx.
Solution
The integrand is a product of two expressions, x and sin x. You
can integrate the second expression, sin x, and differentiating the first
expression, x, makes it simpler. So try integration by parts.
Let f (x) = x and g(x) = sin x.
Then f ! (x) = 1, and an antiderivative of g(x) is G(x) = − cos x.
So the integration by parts formula gives
! !
x sin x dx = f (x)g(x) dx
!
= f (x)G(x) − f ! (x)G(x) dx
!
= x × (− cos x) − 1 × (− cos x) dx
!
= −x cos x + cos x dx
Find the integral in this expression.
= −x cos x + sin x + c.
You can see that in Example 23 the use of the integration by parts formula
changed the problem of integrating x sin x into the problem of integrating
cos x, which is easier. Here are two similar examples for you to try.
Activity 34 Integrating by parts

! !
(a) x cos x dx (b) xex dx
As you become more familiar with integration by parts, you’ll probably

find that you can carry it out more quickly and conveniently if you
remember the informal version of the formula stated below. You can recite
this version in your head as you apply integration by parts, in a similar
way to the informal versions of the product and quotient rules for
differentiation that you met in Unit 7. It’s useful for applying integration
178
by parts in fairly simple cases, like the ones that you’ve seen so far, which
is usually all that you need to do.
Integration by parts formula (informal)

, -
antiderivative
integral of product = (first) ×
of second
,, - , --
derivative antiderivative
− integral of × .
of first of second
In the next example, the integration in Example 23 is repeated, using the

informal version of the integration by parts formula.
Example 24 Integrating by parts, using the informal formula

!
Find the integral x sin x dx.
Solution
The integrand is a product of two expressions, x and sin x. You
can integrate the second expression, sin x, and differentiating the first
expression, x, makes it simpler. So try integration by parts.
Recite the informal version of the formula in your head. As you think
‘first’, write down the first expression, then as you think
‘antiderivative of second’, write down an antiderivative of the second
expression, and so on.
Integrating by parts gives
! !
x sin x dx = x × (− cos x) − 1 × (− cos x) dx
!
= −x cos x + cos x dx
= −x cos x + sin x + c.
You can practise using the informal version of the integration by parts
formula in the next activity. Alternatively, you might prefer to continue
using the formal version for a little longer, and move to the informal
version when you feel more confident with it. You can use the exercises on
integration by parts in the practice quiz and exercise booklet for this unit
for further practice.
179
Activity 35 Integrating by parts

! ! !
(a) x sin(5x) dx (b) xe2x dx (c) xe−x dx
When you use integration by parts, remember that it matters which

expression in the integrand you take to be the ‘first expression’, and which
you take to be the ‘second expression’. If you’re trying to use integration
by parts, but you find that it leads to an integral that’s more complicated
than the original integral, then try swapping the two expressions in the
original integrand.
4.2 Integration by parts in practice

There are several frequently occurring types of integrals that often have
the right characteristics for integration by parts. In this subsection you’ll
meet three of these types. It’s helpful for you to learn to recognise them.
Integrands of the form xn g(x)

You can sometimes find an integral of the form
!
xn g(x) dx,
where n is a positive integer and g(x) is an expression that you can

integrate, by using integration by parts n times.
All the integrals in the previous subsection were of this form, with n = 1,
so you could find them by using integration by parts once. Here’s an
example where you have to use integration by parts twice.
Example 25 Integrating by parts twice

!
Find the integral x2 e4x dx.
Solution
The integral is of the form described above, so try integration by
parts.
180
Integrating by parts gives

! !
x2 e4x dx = x2 × 14 e4x − 2x × 14 e4x dx
!
= 14 x2 e4x − 12 xe4x dx
/
The integral in this expression, xe4x dx, looks easier to find than
the original integral, but it’s still not straightforward to find.
However, it is itself of the form described above, so try integration by
parts again.
, ! -
= 14 x2 e4x − 12 x × 14 e4x − 1 × 14 e4x dx
, ! -
1 2 4x 1 1 4x 1 4x
= 4 x e − 2 4 xe − 4 e dx
Multiply out the brackets. Take care with the minus sign in front
of the brackets – remember that it turns the minus sign inside the
brackets into a plus sign.
!
= 14 x2 e4x − 18 xe4x + 18 e4x dx
Now find the integral in this expression, and simplify the resulting
expression.
= 14 x2 e4x − 18 xe4x + 1
8 × 14 e4x + c
= 14 x2 e4x − 18 xe4x + 1 4x
32 e +c
1 4x 2
= 32 e (8x − 4x + 1) + c.
Activity 36 Integrating by parts twice

! !
2 x
(a) x e dx (b) x2 cos(2x) dx
The method introduced above for dealing with integrands of the form
xn g(x) works just as well if the expression xn is replaced by any
polynomial
/ expression in x of degree n. For example, you can find the
integral (3x2 − 1)e4x dx by integrating by parts twice, and you can find
the integral in the next activity by integrating by parts once.
181
Activity 37 Integrating an expression of the form (ax + b)g(x)

!
Use integration by parts to find the integral (5x + 2)e−5x dx.
Integrands of the form xr ln x

Suppose that you want to integrate an expression of the form
xr ln x,
where r is a constant (any real number). Your first thought, particularly if
r is a positive integer, might be that this is similar to the expressions that
you saw how to integrate earlier in this subsection. However, you can’t
apply the integration by parts formula with ln x as the second expression,
because ln x isn’t an expression that you’ve seen how to integrate.
In fact, you can find an integral of the form above by applying the
integration by parts formula with ln x as the first expression. Here’s an
example.
Example 26 Integrating an expression of the form xn ln x

Find the integral
!
x2 ln x dx.
Solution
The integral is of the form discussed above, so swap the
expressions x2 and ln x, then integrate by parts.
! !
x ln x dx = (ln x) x2 dx
2
!
1 3 1 1 3
= (ln x) × 3 x − × x dx
x 3
!
= 13 x3 ln x − 13 x2 dx
& +
= 13 x3 ln x − 13 13 x3 + c
= 13 x3 ln x − 19 x3 + c
= 19 x3 (3 ln x − 1) + c
182
Activity 38 Integrating expressions of the form xr ln x

! !
3
(a) x ln x dx (b) ln x dx
Hint for part (b): write ln x as (ln x) × 1.
You can use the ideas that you used in Activity 38 to deal with other,
similar integrals that involve the natural logarithm function. Try the
following activity.
Activity 39 Integrating another expression containing the function ln

!
Use integration by parts to find the integral (x − 2) ln(3x) dx.
Integrands of the form eax sin(bx) or eax cos(bx)

You can integrate an expression of the form
eax sin(bx) or eax cos(bx),
where a and b are nonzero numbers, by integrating by parts twice. When
you do this, you obtain an equation that expresses the original integral in
terms of itself, and you can rearrange this equation to find the original
integral. The method is demonstrated in the example below.
When you use this method, you can arrange the exponential expression
and the trigonometric expression in the integrand in either order before
you integrate
/ by parts for/ the first time. For example, you can start with
either eax sin(bx) dx or sin(bx)eax dx. However, you must choose the
same order (exponential before trigonometric, or trigonometric before
exponential) when you integrate by parts for the second time. If you don’t
do that, then the equation that you’ll obtain will reduce to 0 = 0, which
isn’t helpful!
183
Example 27 Finding an integral by expressing it in terms of itself

!
Find the indefinite integral ex cos(2x) dx.
Solution
The integral is of the form discussed above, so use integration by
parts twice. Start by swapping the expressions ex and cos(2x), as this
makes the working slightly easier (it avoids introducing fractions in
the early stages).
We use integration by parts twice.
! !
e cos(2x) dx = cos(2x) ex dx.
x
Integrate by parts.
! !
cos(2x) e dx = cos(2x)e − (−2 sin(2x))ex dx
x x
!
= cos(2x)ex + 2 sin(2x)ex dx
Integrate by parts again.

, ! -
x x x
= cos(2x)e + 2 sin(2x)e − (2 cos(2x))e dx
!
= cos(2x)ex + 2 sin(2x)ex − 4 cos(2x)ex dx
The final term here contains the original integral. Move this term
to the other side of the equation. When you do this, the right-hand
side no longer contains an indefinite integral, so you have to add an
arbitrary constant to that side.
So
! !
x
cos(2x)e dx + 4 cos(2x)ex dx = cos(2x)ex + 2 sin(2x)ex + c;
that is,
!
5 cos(2x)ex dx = cos(2x)ex + 2 sin(2x)ex + c.
To obtain the final answer, divide both sides by 5. (Remember

that you leave the arbitrary constant as c rather than c/5.) Also, it’s
slightly tidier to write the exponential expression before the
trigonometric expression in each term, and to take out a common
factor.
184
This gives
!
ex cos(2x) dx = 15 ex cos(2x) + 25 ex sin(2x) + c
= 15 ex (cos(2x) + 2 sin(2x)) + c.
Activity 40 Finding integrals by expressing them in terms of

themselves
! !
(a) ex sin(3x) dx (b) e2x cos x dx
4.3 Integration by parts for definite integrals

Once you’ve used integration by parts to find an indefinite integral, you
can find any definite integral with the same integrand by using the
fundamental theorem of calculus in the usual way. Here’s an example.
Example 28 Evaluating a definite integral by parts (first method)

! 1
xe7x dx.
0
Solution
Use integration by parts to obtain the indefinite integral of the
integrand.
Integration by parts gives
! !
xe dx = x × 7 e − 1 × 17 e7x dx
7x 1 7x
!
= 7 xe − 7 e7x dx
1 7x 1
= 17 xe7x − 1
7 × 17 e7x + c
1 7x
= 49 e (7x − 1) + c.
185
Apply the fundamental theorem of calculus.

It follows that
! 1 2 11
xe7x dx = 49 1 7x
e (7x − 1)
0 0
2 11
1
= 49 e7x (7x − 1)
0
& 7×1 +
= 49 e (7 × 1 − 1) − e7×0 (7 × 0 − 1)
1
1 7
= 49 (6e − (−1))
1 7
= 49 (6e + 1).
An alternative way to use integration by parts to evaluate a definite

integral is to apply the limits of integration ‘as you go along’. To do this,
you use the following formula, which is just the usual integration by parts
formula with the limits of integration included.
Integration by parts formula for definite integrals

! b ! b
) 'b
f (x)g(x) dx = f (x)G(x) a − f ! (x)G(x) dx.
a a
Here G is an antiderivative of g.
You can remember this formula as an informal version, by applying the

limits of integration a and b in the appropriate way to the informal version
of the usual integration by parts formula.
The next example demonstrates how to apply the limits of integration as
you go along when you integrate by parts. The definite integral in this
example is the same as the one in Example 28.
Example 29 Evaluating a definite integral by parts (second method)

! 1
xe7x dx.
0
Solution
Integrate by parts, applying the limits of integration.
186
! 1 2 11 ! 1
xe7x dx = x × 17 e7x − 1 × 17 e7x dx
0 0 0
! 1
) '1
= 17 xe7x 0 − 17 e7x dx
0
Evaluate the first term, and do the integration in the second

term.
2 11
= 17 (1 × e7×1 − 0 × e7×0 ) − 17 17 e7x
0
) '1
= 17 e7 − 49
1
e7x 0
= 17 e7 − 1
49 (e
7×1
− e7×0 )
1 7 1 7
= 7 e − 49 (e − 1)
1 7 7
= 49 (7e − (e − 1))
1 7
= 49 (6e + 1)
Activity 41 Evaluating a definite integral by parts
Use the integration by parts formula for definite integrals (the method of
Example 29) to evaluate the definite integral
! 1
xe3x dx.
0
Give your answer to four decimal places.
In the next two activities you’re asked to use integration by parts to find
areas.
Activity 42 Using integration by parts to find an area
This question is about the function f (x) = x sin(3x).

(a) Explain why the graph of f lies above the x-axis for values of x in the
interval [π/12, π/6].
(b) Write down a definite integral that gives the area between the graph
of f and the x-axis from x = π/12 to x = π/6.
(c) Use the integration by parts formula for definite integrals to find the
area described in part (b). Give your answer to four decimal places.
187
Activity 43 Using integration by parts to find another area
This question is about the function f (x) = (x − 2)(e−x − 1).

(a) Explain why the graph of f lies above the x-axis for 0 < x < 2 and
below the x-axis for x > 2. You might find it convenient to use a table
of signs as part of your explanation.
(b) Find the total area (not the total signed area) between the graph of f
and the x-axis, from x = 0 to x = 4. Give your answer to four
As mentioned earlier, in some texts the integration by parts formula is

expressed in a different form.
Here’s the formula that you’ve seen in this section:
! !
It involves a function G and its derivative g.

If you denote this pair of functions by g and g ! instead, then you obtain
the alternative version below.
Integration by parts formula (Lagrange notation, alternative

version)
! !
f (x)g ! (x) dx = f (x)g(x) − f ! (x)g(x) dx
As with all calculus formulas, the integration by parts formula can be

stated in Leibniz notation.
If you put u = f (x) and v = g(x) in the formula in the box above, then
you obtain the following form of the formula.
Integration by parts formula (Leibniz notation)

! !
dv du
u dx = uv − v dx
dx dx
188
5 More integration
5 More integration
You’ve now met several integration techniques, and practised using them
on a variety of integrals.
In this section, you’ll start by meeting one further approach, which can be
helpful for integrals that contain trigonometric expressions. Then you’ll
have an opportunity to practise identifying which of the integration
techniques that you’ve met are appropriate for which integrals, for a
variety of different types of integrals.
Finally, you’ll learn how to use a computer for integration.
5.1 Trigonometric integrals

In Section 3 you saw that you can integrate some expressions that contain
trigonometric functions by using integration by substitution. Another
approach that’s often helpful when you want to integrate an expression
that contains trigonometric functions is to start by using one of the
trigonometric identities that you met in Unit 4 to write the expression in a
different form. This can give you an expression that’s easier to integrate.
For example, you can use the identities below to help you integrate the
expressions sin2 x, cos2 x and sin x cos x, as illustrated in the next example.
The first two identities are the half-angle identities, and the third identity
is one of the double-angle identities.
Three trigonometric identities

sin2 θ = 12 (1 − cos(2θ))
cos2 θ = 12 (1 + cos(2θ))
sin(2θ) = 2 sin θ cos θ
189
Example 30 Integrating by using a trigonometric identity

Find the integral
!
sin2 x dx.
Solution
By the half-angle identity for sine,
! !
sin2 x dx = 1
2 (1 − cos(2x)) dx
!
1
= 2 (1 − cos(2x)) dx
= 12 (x − 1
2 sin(2x)) + c
1 1
= 2 x − 4 sin(2x) + c
1
= 4 (2x − sin(2x)) + c.
In the next activity you’re asked to use two of the trigonometric identities
stated near the beginning of this subsection to integrate another two
expressions that contain trigonometric functions.
Activity 44 Integrating by using trigonometric identities

!
(a) Use a half-angle identity to find the integral cos2 θ dθ.
(b) !
Use the double-angle identity in the box above to find the integral
sin x cos x dx.
The trigonometric integral in Example 30 and the one in Activity 44(a)

can’t easily be found without first using a trigonometric identity to write
them in a different form.
However, there’s an alternative way to find the integral in Activity 44(b).
Since the derivative of sin x is cos x, you can use integration by
substitution, taking u = sin x. You’re asked to do that in the next activity.
190
5 More integration
Activity 45 Integrating a trigonometric expression by using

substitution
!
Use integration by substitution to find the integral sin x cos x dx.
In fact there are even two different ways to find the integral in
Activity 44(b), namely
!
sin x cos x dx,
by using integration by substitution. One way is as suggested above: you

use the fact that the derivative of sin x is cos x and hence make the
substitution u = sin x. Alternatively, you can use the fact that the
derivative of cos x is − sin x, write the integral as
!
− (cos x)(− sin x) dx,
and make the substitution u = cos x. The working for each of these
methods is given in the solution to Activity 45.
So you’ve now seen three different ways to integrate the expression
sin x cos x: one way by using a trigonometric identity, and two ways by
using substitution. These three different ways lead to three answers that
look different, as you can see in the solutions to Activities 44(b) and 45.
This illustrates that there can be more than one way to integrate an
expression, and different ways of integrating it can lead to answers that
look different, though of course these answers will be equivalent.
If an answer that you obtain for an integral is different from the answer
given in a book or by a computer, but you think that you haven’t made a
mistake, then you may be able to check that the two answers are equivalent
by using algebraic manipulation. You may need to use trigonometric
identities – there are examples of this in the solution to Activity 45.
A rougher, but sometimes quicker, check is to plot the graphs of the two
different answers on a computer (taking the arbitrary constants to be zero,
for example), and check that they seem to be vertical translations of each
other. This shows that the two different answers seem to be equivalent
once the arbitrary constants are taken into account. Again, there’s an
example of this in the solution to Activity 45.
In this subsection you’ve seen just a few examples of how you can use
trigonometric identities and integration by substitution to help you
integrate expressions that contain trigonometric functions. There are many
more possibilities, some of which are covered in the module Essential
mathematics 2 (MST125).
191
5.2 Choosing a method of integration

When you have an integral that you want to find, it may not be obvious
which method to try. Also, sometimes you might need to combine two or
more techniques – for example you might be able to use algebraic
manipulation, the constant multiple rule and the sum rule to express an
integral in terms of simpler integrals, then use integration by substitution
or integration by parts to find the simpler integrals.
Here are some things to think about when you have an integral that you
want to find.
Choosing a method for finding an integral

• Is it a standard integral? Consult the table in the Handbook.
• Can you rearrange the integral to express it as a sum of constant
multiples of simpler integrals?
• Is the integrand of the form f (ax) or f (ax + b), where a and b
are constants and f is a function that you can integrate? If so,
use the rule for integrating a function of a linear expression,
or use integration by substitution, with u = ax or u = ax + b as
appropriate.
• Can the integrand be written in the form
where f is a function that you can integrate? If so, use
integration by substitution. Start by setting the ‘something’
equal to u.
• Is the integrand of the form
f (x)g(x),
where f is a function that becomes simpler when differentiated,
and g is a function that you can integrate? (For example, f (x)
might be x or x2 or ln x.) If so, try integration by parts.
• If the integrand contains trigonometric functions, can you use
trigonometric identities to rewrite it in a form that’s easier to
integrate?
When you’re finding an indefinite integral, try to always remember to add

the arbitrary constant ‘ + c’. It’s easy to forget it!
In the next activity you can practise identifying which method of
integration to use.
192
5 More integration
Activity 46 Choosing a method of integration
For each integral below, suggest a method of integration to use. Where you
suggest substitution, also suggest a suitable equation for u in terms of x.
You are not asked to actually find the integrals, but solutions are included
so you can do so if you wish.
! ! !
(a) x3 cos(9x4 ) dx (b) x cos(5x) dx (c) (x + cos(5x)) dx
! ! !
1 x2
(d) x(x2 + 3) dx (e) dx (f) dx
1 + 3x2 (7 − x3 )7
! ! !
−x/3
&1 + &1 +
(g) xe dx (h) sin 2 x cos 2 x dx (i) (x − 1)4 dx
! ! !
x x 2
(j) e (1 − 2e ) dx (k) x(1 + sin(x )) dx (l) x(3x2 − 2)8 dx
You’ll learn further useful strategies for integration in the module Essential
mathematics 2 (MST125), if it’s in your study programme. In particular,
you’ll learn more about integration by substitution and integration by
using trigonometric identities, and you’ll study partial fractions, which are
useful when you want to integrate rational functions.
193
5.3 Integration using a computer

There are many situations in which it’s helpful to use a computer for
integration, such as when you want to integrate an expression that’s
difficult to integrate by hand. In the first activity of this subsection you
can learn how to use the module computer algebra system to find
indefinite integrals and evaluate definite integrals.
Activity 47 Using a computer for integration
Work through Section 9 of the Computer algebra guide.
You can use the skills that you learned in Activity 47 in the next activity.
Activity 48 Using a computer to work with a complicated function
This activity is about the function

x2 − 10x + 15
f (x) = √ .
1+ x+1
Do parts (b) to (d) below by using the module computer algebra system.
Give your answers to parts (c) and (d) to three significant figures.
(a) What is the domain of f ?
(b) Plot the graph of f .
(c) Find the two x-intercepts of f .
(d) Find the area between the graph of f and the x-axis, between the two
x-intercepts.
The next two activities are the final ones of the unit. In them, you’re
asked to use your knowledge of integration to solve practical problems.
The definite integrals involved in the problems can’t easily be evaluated by
using the fundamental
√ theorem of calculus, as it’s difficult to find
3 2
antiderivatives of x + 1 and sin(t /150). However, you can use a
computer to find approximate values.
194
5 More integration
Activity 49 Using integration to solve a practical problem
Consider the plastic edging shown below. The shape of its cross-section is
given by the equation
"
y = 13 x3 + 1 (−1 ≤ x ≤ 1.5),
where x and y are measured in centimetres.
Calculate the volume of plastic that is required to make one metre of the
edging. Give your answer in cubic centimetres to three significant figures.
Activity 50 Using integration to solve another practical problem
Suppose that an object moves in a straight line, and its velocity v

(in m s−1 ) at time t (in seconds) is given by the equation
, 2 -
t
v = sin (0 ≤ t ≤ 20),
150
as shown in Figure 42.
If the displacement of the object from a particular reference point on the
line at time t = 0 is 6 m, as shown below, what is the displacement of the
object from this reference point at time t = 10? Give your answer to the
nearest centimetre.

v = sin(x2 /150) (0 ≤ t ≤ 20)
195
You’ve now come to the end of the introduction to basic calculus in this
module. You can use the skills that you’ve acquired to solve many different
types of problems in mathematics, engineering and science, for example.
These skills also equip you to study many further topics in these areas.
You’ll make a start on that in Unit 11, in which you’ll use calculus in your
study of Taylor polynomials. If your study programme includes further
mathematics modules, then in some of these you may study more
advanced calculus, with even more powerful applications.
Learning outcomes
After studying this unit, you should be able to:
• understand what is meant by a definite integral
• understand how to use subintervals to find an approximate value for a
definite integral
• use integral notation
• understand the fundamental theorem of calculus
• evaluate some definite integrals by using the fundamental theorem of
calculus
• integrate by substitution
• integrate by parts
• appreciate how to use trigonometric identities and integration by
substitution to help you integrate some trigonometric expressions
• identify a suitable method of integration for some types of integrals
• use a computer for integration.
196
Solutions to activities
Solution to Activity 1 So the heights of the rectangles are:
If we divide the interval [−18, 18] into six f (−3) = 3 − (−3)2 = −6
subintervals of equal width, then each subinterval f (−2) = 3 − (−2)2 = −1
has width 36/6 = 6.
f (−1) = 3 − (−1)2 = 2
The left endpoints of the six subintervals are
f (0) = 3 − 02 = 3
− 18, −18 + 6, −18 + 2 × 6,
f (1) = 3 − 12 = 2
− 18 + 3 × 6, −18 + 4 × 6, −18 + 5 × 6
f (2) = 3 − 22 = −1.
that is
So we obtain the following approximate value for
−18, −12, −6, 0, 6, 12. the signed area:
So the heights of the rectangles are: (−6 × 1) + (−1 × 1) + (2 × 1)
1
f (−18) = 9 − 36 × (−18)2 = 0 + (3 × 1) + (2 × 1) + (−1 × 1) = −1.
1 2
f (−12) = 9 − 36 × (−12) =5
Solution to Activity 4
f (−6) = 9 − 36 1
× (−6)2 = 8
The exact value of the signed area between the
1
f (0) = 9 − 36 × 02 = 9 graph of the function f (x) = 3 − x2 and the x-axis
1
f (6) = 9 − 36 × 62 = 8 from x = −3 to x = 3 is 0.
f (12) = 9 − 361
× 122 = 5.
So we obtain the following approximate value for
the area: (a) The signed area from x = 2 to x = 6 is 4.0.
(0 × 6) + (5 × 6) + (8 × 6) (b) The signed area from x = 2 to x = 6 is 4.0,
+ (9 × 6) + (8 × 6) + (5 × 6) = 210. so the signed area from x = 6 to x = 2 is −4.0.
(c) The signed area from x = 2 to x = 10 is
Solution to Activity 2 4.0 − 5.4 = −1.4.
(a) The signed area from x = 1 to x = 5 is −6.3. (d) From part (c), the signed area from x = 2 to
(b) The signed area from x = 5 to x = 7 is 1.3. x = 10 is −1.4, so the signed area from x = 10
to x = 2 is 1.4.
(c) The signed area from x = 1 to x = 7 is
−6.3 + 1.3 = −5. (e) The signed area from x = 8 to x = 8 is 0.
(d) The signed area from x = 1 to x = 9 is (f) The signed area from x = −3 to x = 2 is
−6.3 + 1.3 − 5.9 = −10.9. 4.1 − 2.4 = 1.7, so the signed area from x = 2 to
x = −3 is −1.7.
Solution to Activity 3 (g) The signed area from x = −1 to x = 10 is
If we divide the interval [−3, 3] into six subintervals −2.4 + 4.0 − 5.4 = −3.8, so the signed area
of equal width, then each subinterval has width from x = 10 to x = −1 is 3.8.
6/6 = 1.
The left endpoints of the six subintervals are
−3, −2, −1, 0, 1, 2.
197
Solution to Activity 6 Solution to Activity 8

! −3 )1 '
2 5 1 1
(a) 2x 3 = × 52 − × 32 = 12 (25 − 9) = 8.
(a) f (x) dx = −5. 2 2
−5 ) '2π
! (b) cos x 0 = cos(2π) − cos 0 = 1 − 1 = 0.
2
(b) f (x) dx = 6. ) '1
(c) ex −1 = e1 − e−1 = e − (1/e) = 2.35 (to 3 s.f.).
−3
! 7
(c) f (x) dx = −11. Solution to Activity 9
!
2 (a) An antiderivative of x2 is 13 x3 , so
2 ! 1
(d) f (x) dx = −5 + 6 = 1. ) '1
−5
x2 dx = 13 x3 −1
! −1
7
(e) f (x) dx = 6 − 11 = −5. = 1
3 × 13 − 1
3 × (−1)3
2
−3 = 3.
! 7
(f) f (x) dx = −5 + 6 − 11 = −10. (b) An antiderivative of et is et , so
! 2
−5 ) '2
! 5 et dt = et 0
(g) f (x) dx = 0. 0
5 = e2 − e 0
! −3 ! 2
= e2 − 1 = 6.39 (to 3 s.f.).
(h) f (x) dx = − f (x) dx = −6.
2 −3 (c) An antiderivative of 1/u (for u > 0) is ln u, so
! 2 ! 7 ! 4
1 ) '4
(i) f (x) dx = − f (x) dx = −(−11) = 11. du = ln u 1
7 2 1 u
! −5 ! −3 = ln 4 − ln 1
(j) f (x) dx = − f (x) dx = −(−5) = 5.
−3 −5
= ln 4 = 1.39 (to 3 s.f.).
! −3 ! 7 (d) An antiderivative of sec2 θ is tan θ, so
(k) f (x) dx = − f (x) dx ! π/4
) 'π/4
7 −3
sec2 θ dθ = tan θ −π/4
= −(−5) = 5, −π/4
π . π#
by part (e). = tan − tan −
! −5 ! 7 4 4
(l) f (x) dx = − f (x) dx = 1 − (−1)
7 −5 = 2.
= −(−10) = 10,
(e) An antiderivative of 1/(1 + u2 ) is tan−1 u, so
by part (f). ! 1
1 ) '1
du = tan−1 u −1
Solution to Activity 7 −1 1 + u
2
! 9 ! 7 ! 9
= tan−1 (1) − tan−1 (−1)
f (x) dx = f (x) dx + f (x) dx π . π#
2 2 7 = − −
so 4 4
! 9
= π/2
−15 = −11 + f (x) dx
7 = 1.57 (to 3 s.f.).
and hence
! 9
f (x) dx = −15 − (−11) = −4.
7
198
! 1
! ! (b) (u − 2u2 ) du
4 √ 4
−1/2
(a) 3 x dx = 3x1/2 dx 2 11
1 2
1
(
1
04 = − 2 × 13 u3
2u
−1/2
1 3/2 2 11
= 3× x
3/2 1 = 12 u2 − 23 u3
2 14 −1/2
= 2x3/2 2 11 2 11
1 = 12 u2 − 23 u3
−1/2 −1/2
= 2 × 4 − 2 × 13/2
3/2
2 11 2 11
=2×8−2 = 12 u2 − 23 u3
−1/2 −1/2
= 14. & 2 + 2& 3 +
! = 2 1 − (− 2 ) − 3 1 − (− 12 )3
1 1 2
0
(b) x(1 + x) dx = 12 (1 − 14 ) − 23 (1 + 18 )
−1 1 3 2 9
! 0 = 2 × 4 − 3 × 8
= (x + x2 ) dx = 3
8 − 3
4
−1
)1 2 1 3
'0 = − 38 .
= 2x + 3x −1
= ( 12 ×0 + 2 1
× 03 ) − ( 12 (−1)2 + 13 (−1)3 ) Solution to Activity 12
3 √ √
! 7 ! 7
= −( 12 − 13 ) 1 3
(a) √ 2x dx = 1
2 √ x3 dx
= − 16 . 3 3
2 1√7
! 4 ! 4, - 1 1 4
r+5 5 = 2 4x √
(c) dr = dr 1+ 3
2 r 2 r ) '√
) '4 1 4 √ 7
= 8 x
= r + 5 ln r 2 .√ 3
√ #
= (4 + 5 ln 4) − (2 + 5 ln 2) = 1
8 ( 7)4 − ( 3)4
= 2 + 5(ln 4 − ln 2) = 18 (72 − 32 )
= 2 + 5 ln(4/2) = 18 (49 − 9)
= 2 + 5 ln 2. = 5.
! π/4
Solution to Activity 11 (b) (sin x + cos x) dx
! 5 ( 05 −π/4
1 5/2 ! !
(a) x3/2 dx = x π/4 π/4
3 5/2 3 = sin x dx + cos x dx
2 15 −π/4 −π/4
2 5/2
= 5x ) 'π/4 ) 'π/4
3 = − cos x −π/4 + sin x −π/4
2 15
= 25 x5/2 ) 'π/4 ) 'π/4
3 = − cos x −π/4 + sin x −π/4
= 2 5/2
− 35/2 ) . .π# . π ##
5 (5
= − cos − cos −
= 16.1 (to 3 s.f.). . π #4 . π #4
+ sin − sin −
, 4 - 4 , -
1 1 1 1
=− √ −√ + √ − −√
2 2 2 2
2
=√
2
√
= 2.
199
Solution to Activity 13 Hence the shaded area is 21 333 cm2 , to the

The required area is given by nearest cm2 .
! 18 . #
1 2
9 − 36 x dx Solution to Activity 15
−18
2 118 (a) (i) From t = 0 to t = 1 the displacement of
1 1 3 the object changes by 2.3 cm.
= 9x − 36 × 3x
−18
2 118 (ii) From t = 3 to t = 4 the displacement of
1 3
= 9x − 108 x the object changes by −1.8 cm. That is, it
−18
)'18 2 118 changes by 1.8 cm in the negative direction.
= 9x −18 − 108 1
x3
−18 (iii) From t = 0 to t = 3 the displacement of
) '18 1
) 3 '18 the object changes by
= 9 x −18 − 108 x −18
3 (2.3 + 5.9 + 2.7) cm = 10.9 cm.
= 9(18 − (−18)) − 1
108 (18 − (−18)3 )
1 3 3 (iv) From t = 3 to t = 6 the displacement of
= 9 × 36 − 108 (18 − (−18 )) the object changes by
1 3
= 9 × 36 − 108 (2 × 18 ) (−1.8 − 2.0 − 0.3) cm = −4.1 cm.
= 324 − 108 That is, it changes by 4.1 cm in the
= 216. negative direction.
So the area of the cross-section of the roof is 216 m2 . (b) (i) The displacement of the object from its
starting position at time t = 2 is
(2.3 + 5.9) cm = 8.2 cm.
(a) The x-intercepts are given by
1 2 (ii) By (a) parts (iii) and (iv), the
32 x − 200 = 0 displacement of the object from its
1 2
32 x = 200 starting position at time t = 6 is
2
x = 6400 (10.9 − 4.1) cm = 6.8 cm.
√
x = ± 6400 (c) By parts (a)(iii) and (a)(iv), the object travels
x = ±80. 10.9 cm in the positive direction, then 4.1 cm in
So the x-intercepts are −80 and 80. the negative direction, so the total distance that
it travels is
(b) The shaded area is
! 80 . # (10.9 + 4.1) cm = 15 cm.
1 2
x − 200 dx
32
−80 Solution to Activity 16
2 180
= 32 1
× 13 x3 − 200x (a) The change in the displacement of the object
−80 from time t = 0 to time t = 1 is given by
2 180 ! 1 2 11
1 3
= 96 x − 200x (3 − 32 t)dt = 3t − 32 × 12 t2
−80 0
1
) 3 '80 ) '80 0
2 11
= 96 x −80 − 200 x −80
= 3t − 34 t2
1 3 3
= 96 (80 − (−80) ) − 200(80 − (−80)) 0
) '1 3 ) 2 '1
= 1 3 =3 t 0−4 t 0
96 (2 × 80 ) − 200 × 160
80 3 = 3(1 − 0) − 34 (12 − 02 )
= − 32 000 3
48 =3− 4
64 000 = 94 = 2.25.
=−
3
That is, the object travels 2.25 m in that time
= −21 333 13 . (in the positive direction).
200
(b) The change in the displacement of the object Solution to Activity 19

from time t = 4 to time t = 5 is given by !
! 5 2 15 (a) ex dx = ex + c.
(3 − 32 t)dt = 3t − 32 × 12 t2 ! !
4 4
2 15 (b) (x + 3)(x − 2) dx = (x2 + x − 6) dx
= 3t − 34 t2
4
) '5 3 ) 2 '5 = 13 x3 + 12 x2 − 6x + c.
=3 t 4−4 t 4 ! !
1
= 3(5 − 4) − 34 (52 − 42 ) (c) dx = x−3 dx
x3
=3− 3
×9 1 −2
4 = x +c
= − 15 −2
4
1
= −3.75. = − 2 + c.
2x
That is, the object travels 3.75 m in that time !
(in the negative direction). (d) (sin θ − cos θ) dθ = − cos θ − sin θ + c.
!
Solution to Activity 17 1
(e) dx = tan−1 x + c.
The change in the displacement of the car during 1 + x2
!
those four seconds is given by −3
! 4 2 14 (f) du = −3 tan−1 u + c.
1 + u2
(28 − 7t) dt = 28t − 72 t2 !
0 0 1
) '4 ) '4 (g) dr = 12 tan−1 r + c.
= 28 t 0 − 72 t2 0 2(1 + r2 )
!
= 28(4 − 0) − 72 (42 − 0) (1 + x2 )
(h) dx = 12 (x + 13 x3 ) + c
= 112 − 56 2
= 56. = 16 (3x + x3 ) + c.
That is, the car travels 56 m during those four
seconds.
(In fact, since the required answer is the area of the ! , - ! !
shaded triangle in Figure 39, you can calculate it in 3 1 3 1
(a) − dx = dx − dx
the following more straightforward way. x 1 + x2 x 1 + x2
! !
The change in the displacement of the car during 1 1
=3 dx − dx
the four seconds is x 1 + x2
( 12 × 4 × 28) m = 56 m.) = 3 ln |x| − tan−1 x + c.
!
Solution to Activity 18 (b) (cosec x)(cosec x + cot x) dx
The change in the displacement of the car is the !
area shaded in Figure 40, which is given by = (cosec2 x + cosec x cot x) dx
1 1 1 ! !
2 × 5 × 15 − 2 × 4 × 12 = 2 (75 − 48) 2
= cosec x dx + cosec x cot x dx
= 12 × 27 = 13.5.
So the car travels 13.5 m during the fifth second. = − cot x − cosec x + c.
(There are various ways to work out the area
shaded in Figure 40. It is worked out above by
subtracting the area of one triangle from the area of
another, but alternatively you could add the areas
of a rectangle and a triangle, or use the formula for
the area of a trapezium.)
201
Solution to Activity 21 du
(f) Let u = sin x; then = cos x. So
du ! , - dx ! , -
(a) Let u = cos x; then = − sin x. So 1 1
! dx ! cos x dx = du
1 + sin2 x 1 + u2
ecos x (− sin x) dx = eu du
= tan−1 u + c
= eu + c = tan−1 (sin x) + c.
= ecos x + c.
du Solution to Activity 22
3
(b) Let u = x ; then = 3x2 . So du
! dx ! (a) Let u = 4 + cos x; then = − sin x. So
& 3
+ 2 ! dx !
sin(x ) (3x ) dx = sin(u) du
(4 + cos x) (− sin x) dx = u7 du
7
= − cos u + c
= − cos(x3 ) + c. = 18 u8 + c
du = 18 (4 + cos x)8 + c.
(c) Let u = sin x; then = cos x. So
! , - dx ! du
1 1 (b) Let u = 1 + x2 ; then = 2x. So
cos x dx = du ! " dx !
sin x u √
1 + x2 (2x) dx = u du
= ln |u| + c
!
= ln | sin x| + c. = u1/2 du
du
(d) Let u = cos x; then = − sin x. So 1 3/2
! , - dx ! = u +c
1 1 3/2
(− sin x) dx = du
cos x
2 u2 = 23 (1 + x2 )3/2 + c.
!
du
= u−2 du (c) Let u = x5 − 8; then = 5x4 . So
! dx !
1 −1
=u +c (x − 8) (5x ) dx = u10 du
5 10 4
−1
1 = 1 11
=− +c 11 u +c
u 5
1 = 1
11 (x − 8)11 + c.
=− +c du
cos x (d) Let u = ex + 5; then = ex . So
= − sec x + c. ! , - dx!
1 1
(A quicker way to find this indefinite integral, x
e dx = du
x
e +5 u
without using substitution, is to write the
integrand as − sec x tan x and use the table of = ln |u| + c
standard indefinite integrals.) = ln |ex + 5| + c
du = ln(ex + 5) + c.
(e) Let u = sin x; then = cos x. So
! dx! du
(e) Let u = 5 + 2x3 ; then = 6x2 . So
sin4 x cos x dx = u4 du ! dx !
& 3
+ 2
sin(5 + 2x ) (6x ) dx = sin u du
= 15 u5 + c
= 1
sin5 x + c. = − cos u + c
5
= − cos(5 + 2x3 ) + c.
202
Solution to Activity 23 du
(b) Let u = 1 + x2 ; then = 2x. So
du ! " !dx." #
(a) Let u = ex ; then = ex . So
! dx ! x 1 + x2 dx = 1 + x2 x dx
x x
(cos(e )) e dx = cos u du ! ." #
= 12 1 + x2 × 2x dx
= sin u + c !
√
= sin(ex ) + c. = 12 u du
du !
(b) Let u = 1 + sin x; then = cos x. So
! dx
! = 12 u1/2 du
e1+sin x (cos x) dx = eu du = 1 2 3/2
2 × 3u +c
= eu + c = 1 3/2
3u +c
= e1+sin x + c. = 1 2 3/2
3 (1 + x ) + c.
du du
(c) Let u = x10 + 6; then = 10x9 . So (c) Let u = cos x; then = − sin x. So
! , - dx ! dx
1 1 ! !
9
10
(10x ) dx = du cos x sin x dx = − (cos4 x)(− sin x) dx
4
x +6 u
= ln |u| + c !
= − u4 du
= ln |x10 + 6| + c
= ln(x10 + 6) + c. = − 15 u5 + c
du = − 15 cos5 x + c.
(d) Let u = cos x; then = − sin x. So
! dx !
(cos3 x)(− sin x) dx = u3 du
du
1 4 (a) Let u = 1 + 2x2 ; then = 4x. So
= 4u + c ! ! ,dx -
= 1 4 x 1
4 cos x + c. dx = x dx
1 + 2x2 1 + 2x2
! , -
Solution to Activity 24 1 1
=4 (4x) dx
du 1 + 2x2
(a) Let u = x3 ; then = 3x2 . So !
! dx ! 1
& + = 14 du
2 3
x cos(x ) dx = cos(x3 ) x2 dx u
! = 14 ln |u| + c
= 3 (cos(x3 ))(3x2 ) dx
1
= 1
4 ln |1 + 2x2 | + c
! = 1
ln(1 + 2x2 ) + c
1 4
= 3 cos u du (since 1 + 2x2 is always positive).
1
= 3 sin u + c
= 1
3 sin(x3 ) + c.
203
du du
(b) Let u = 2 + ex ; then = ex . So (c) Let u = 1 + 12 sin x; then = 12 cos x. So
! dx
! , - ! dx
ex 1
dx = ex dx (1 + 12 sin x)2 cos x dx
(2 + ex )2 (2 + ex )2 !
!
1 = 2 (1 + 12 sin x)2 ( 12 cos x) dx
= du
u2 !
!
= u−2 du = 2 u2 du
= −u−1 + c = 2 × 13 u3 + c
1 = 23 u3 + c
=− +c
u
1 = 23 (1 + 12 sin x)3 + c.
=− + c.
2 + ex Solution to Activity 27
du du
(c) Let u = 3 − sin x; then = − cos x. So Let u = cos x; then = − sin x. So
! dx
! , - dx
cos x 1 ! !
dx = − (− cos x) dx sin x
3 − sin x 3 − sin x tan x dx = dx
! cos x
! , -
1 1
=− du = (sin x) dx
u cos x
= − ln |u| + c ! , -
1
= − ln |3 − sin x| + c =− (− sin x) dx
cos x
!
= − ln(3 − sin x) + c 1
=− du
(since 3 − sin x is always positive). u
= − ln |u| + c
Solution to Activity 26 = − ln | cos x| + c.
du
(a) Let u = cos x; then = − sin x. So
! dx ! Solution to Activity 28
ecos x sin x dx = − ecos x (− sin x) dx du
(a) Let u = 6x − 1; then = 6. So
! ! ! dx
=− eu du e6x−1 dx = 1
6 e6x−1 × 6 dx
!
= −eu + c
= 1
eu du
= −ecos x + c. 6
du = 16 eu + c
(b) Let u = 3 − ex ; then = −ex . So
! dx
! , - = 16 e6x−1 + c.
ex 1
dx = − (−ex ) dx du
3 − ex 3 − ex (b) Let u = 4x; then = 4. So
! ! dx!
1
=− du sin(4x) dx = 14 (sin(4x)) × 4 dx
u
= − ln |u| + c !
1
= − ln |3 − ex | + c. = 4 sin u du
= 14 (− cos u) + c
= − 14 cos u + c
= − 14 cos(4x) + c.
204
du du
(c) Let u = −9x; then = −9. So (g) Let u = 2x + 1; then = 2. So
! !dx ! dx
! , -
1 1
e−9x dx = − 19 e−9x × (−9) dx dx = 1
2 × 2 dx
! (2x + 1)2 (2x + 1)2
!
= − 19 eu du 1 1
=2 du
u2
!
= − 19 eu + c
=2 1
u−2 du
= − 19 e−9x + c.
du = − 12 u−1 + c
(d) Let u = −x/3; then = − 13 . So 1 1
! !dx =− × +c
2 u
e−x/3 dx = −3 e−x/3 × (− 13 ) dx 1
! =− + c.
2(2x + 1)
u
= −3 e du
du
(h) Let u = 1 − x; then = −1. So
= −3eu + c ! dx
! , -
= −3e−x/3 + c. 1 1
dx = − × (−1) dx
du (1 − x)3 (1 − x)3
(e) Let u = 3 − 7x; then = −7. So !
dx 1
! ! =− 3
du
u
1
cos(3 − 7x) dx = − 7 (cos(3 − 7x)) × (−7) dx !
! = − u−3 du
= − 17 cos u du
= −(− 12 u−2 ) + c
= − 17 sin u + c 1
= 2 +c
1
= − 7 sin(3 − 7x) + c. 2u
1
du = + c.
(f) Let u = 4 − x; then = −1. So 2(1 − x)2
! ! dx
, - du
1 1 (i) Let u = 6x; then = 6. So
dx = − × (−1) dx dx !
4−x 4−x !
!
1 sec2 (6x) dx = 16 (sec2 (6x)) × 6 dx
=− du !
u
= − ln |u| + c = 6 sec2 u du
1
= − ln |4 − x| + c. = 1
tan u + c
6
1
= 6 tan(6x) + c.
205
!
Solution to Activity 29 (c) e3t+4 dt = 13 e3t+4 + c.
du !
(a) Let u = 2x; then = 2. So
! dx ! (d) cos( 12 θ) dθ = 2 sin( 12 θ) + c.
1 1
√ dx = " dx !
1 − 4x2 1 − (2x)2
! (e) e−4p dp = − 14 e−4p + c.
1 1
=2 " × 2 dx !
1 − (2x)2 1
! (f) dx = 17 ln |7x − 4| + c.
1 7x − 4
= 12 √ du !
1 − u2 1
(g) dr = − ln |1 − r| + c.
= 12 sin−1 u + c 1−r
!
= sin−1 (2x) + c.
1
2 (h) sec2 (5x − 1) dx = 15 tan(5x − 1) + c.
√ du √ !
(b) Let u = 2 x; then = 2. So
! dx ! (i) sin( 15 x) dx = −5 cos( 15 x) + c.
1 1 1 √
dx = √ &√ + × 2 dx !
1 + 2x2 2
2 1+ 2x (j) e−x dx = −e−x + c.
!
1 1 !
=√ du
2 1 + u2 (k) ex/2 dx = 2ex/2 + c.
1 !
= √ tan−1 u + c
2 (l) sec2 (2 − 3p) dp = − 13 tan(2 − 3p) + c
1 .√ #
= √ tan−1 2 x + c. = 13 tan(3p − 2) + c.
2
du (The final expression was obtained by using the
(c) Let u = 32 x; then = 32 . So trigonometric identity tan(−θ) = − tan θ.)
! dx ! ! , - !
1 1 1 2−x & +
dx = 4 dx (m) sin dx = sin 23 − 13 x dx
4 + 9x2 1 + 94 x2 3
! & & ++
1 = −3 − cos 23 − 13 x + c
= 14 & 3 +2 dx , -
1 + 2x 2−x
! = 3 cos + c.
1 2 1 3 3
=4×3 & 3 +2 × 2 dx
1 + 2x Solution to Activity 31
!
1
= 16 3
& 3 +2 × 2 dx (a) The velocity v (in m s−1 ) of the object at time t
1 + 2x (in seconds) is given by
! & +
1 1 v = 2 cos 12 t + 5 .
=6 du
1 + u2 Hence the displacement s (in m) of the object
= 16 tan−1 u + c at time t (in seconds) is given by
& + !
= 16 tan−1 32 x + c. & +
s = 2 cos 12 t + 5 dt
Solution to Activity 30 & +
! = 2 × 2 sin 12 t + 5 + c
& +
(a) 1
cos(10x) dx = 10 sin(10x) + c. = 4 sin 12 t + 5 + c,
! where c is an arbitrary constant.
(b) sin(2 − 5x) dx = − 15 (− cos(2 − 5x)) + c
1
= 5 cos(2 − 5x) + c.
206
When t = 0, s = 4. Hence du
& + (b) Let u = cos x; then = − sin x. Putting
4 = 4 sin 12 × 0 + 5 + c dx
x = π/2 gives u = 0 and putting x = π gives
4 = 4 sin(5) + c u = −1. So
! π ! π
c = 4 − 4 sin(5) 3
cos x sin x dx = − cos3 x(− sin x) dx
c = 7.835 . . . π/2 π/2
! −1
c = 7.84 (to 2 d.p.).
=− u3 du
So the equation for the displacement of the 0
object in terms of time is 2 1−1
& + = − 14 u4
s = 4 sin 12 t + 5 + 7.84. 0
1
) 4 '−1
(b) When t = 10, = −4 u 0
& + & +
s = 4 sin 12 × 10 + 5 + 7.835 . . . = − 14 (−1)4 − 04
= 4 sin(10) + 7.835 . . . = − 14
= 5.7 (to 2 s.f.). = −0.25.
So the displacement of the object at time (Notice that in the second line of the
10 seconds is 5.7 m (to 2 s.f.). manipulation above we obtain an integral,
/ −1 3
0
u , in which the lower limit of integration is
Solution to Activity 32 greater than the upper limit of integration. An
du alternative way to proceed from this line is to
(a) Let u = 1 + 2x2 ; then = 4x.
dx use the fact that, in general,
/a /b
Putting x = 0 gives u = 1 and putting x = 1 b
f (x) dx = − a f (x) dx. Then the
gives u = 3. So manipulation continues like this:
! 1 ! 1, - ! −1 ! 0
x 1
2
1
dx = 4 (4x) dx u3 du = u3 du
0 1 + 2x 0 1 + 2x2 0 −1
! 3 2 10
1 = 1 4
4u
1
=4 du
1 u
−1
) '3 1
) '
4 0
= 14 ln |u| 1 = 4 u −1
& 4 +
= 14 (ln 3 − ln 1) = 1
4 0 − (−1)4
= 14 ln 3 = − 14
= 0.275 (to 3 s.f.). = −0.25.)
du
(c) Let u = −3x2 ; then = −6x.
dx
Putting x = −1 gives u = −3 and putting x = 0
gives u = 0. So
! 0 ! 0. #
−3x2 2
xe dx = − 61
e−3x (−6x) dx
−1 −1
!0
= − 16 eu du
−3
) '0
= − 16 eu −3
& +
= − 16 e0 − e−3
& +
= 16 e−3 − 1
= −0.158 (to 3 s.f.).
207
Solution to Activity 33 Hence

! !
du x
xe dx = f (x)g(x) dx
(a) Let u = 1 + x; then = 1. Also, x = u − 1. So
! !dx !
√ √
x 1 + x dx = (u − 1) u du = f (x)G(x) − f " (x)G(x) dx
! . # !
= u3/2 − u1/2 du = xe − ex dx
x
u5/2 u3/2 = xex − ex + c

= − +c = ex (x − 1) + c.
5/2 3/2
= 25 (1 + x)5/2 − 23 (1 + x)3/2 + c. Solution to Activity 35
!
du & +
(b) Let u = x + 3; then = 1. Also, x = u − 3, so (a) x sin(5x) dx = x − 15 cos(5x)
dx
x −! 2 = u − 5. Hence !
! & +
x−2 u−5 − 1 × − 15 cos(5x) dx
dx = du
(x + 3)3 u3 !
! 1 1
= − x cos(5x) + cos(5x) dx
= (u−2 − 5u−3 ) du 5 5
= − 15 x cos(5x) + 15 × 15 sin(5x) + c
= −u−1 + 52 u−2 + c
1 5 = − 15 x cos(5x) + 25 1
sin(5x) + c
=− + 2 +c = 251
sin(5x) − 15 x cos(5x) + c.
u 2u
1 5 ! !
=− + + c. 2x
& 1 2x + & +
x + 3 2(x + 3)2 (b) xe dx = x 2 e − 1 × 12 e2x dx
!
= 2 xe − 2 e2x dx
1 2x 1
(a) Let f (x) = x and g(x) = cos x. Then f " (x) = 1 = 12 xe2x − 1 1 2x
2 × 2e +c
and an antiderivative of g(x) is G(x) = sin x. = 1
2 xe
2x
− 1 2x
4e +c
Hence
! ! 1 2x
= 4 e (2x −1) + c.
x cos x dx = f (x)g(x) dx ! !
& + & +
! (c) xe−x dx = x −e−x − 1 × −e−x dx
= f (x)G(x) − f " (x)G(x) dx !
! = −xe−x + e−x dx
= x sin x − sin x dx
= −xe−x + (−e−x ) + c
= x sin x − (− cos x) + c = −xe−x − e−x + c
= x sin x + cos x + c. = −e−x (x + 1) + c.
(b) Let f (x) = x and g(x) = ex . Then f " (x) = 1
and an antiderivative of g(x) is G(x) = ex .
208
Solution to Activity 36 Solution to Activity 38

! ! ! !
2 x 2 x x 3
(a) x e dx = x e − 2xe dx (a) x ln x dx = (ln x)x3 dx
! !
1 1 4
= x2 ex − 2 xex dx = (ln x) × 1 4
4x − × x dx
x 4
, ! - !
2 x x x
= x e − 2 xe − 1 × e dx = 14 x4 ln x − 1
4 x3 dx
!
= 14 x4 ln x − 1
× 14 x4 + c
= x e − 2xe + 2 ex dx
2 x x 4
1 4
= 16 x (4 ln x − 1) + c.
2 x x
= x e − 2xe + 2e + c x ! !
= ex (x2 − 2x + 2) + c. (b) ln x dx = (ln x) × 1 dx
! !
1
(b) x2 cos(2x) dx = (ln x)x − × x dx
x
! !
= x2 ( 12
sin(2x)) − 2x( 12 sin(2x)) dx = x ln x − 1 dx
!
1 2 = x ln x − x + c
= 2 x sin(2x) − x sin(2x) dx
= x(ln x − 1) + c.
= 12 x2 sin(2x)
, ! - Solution to Activity 39
1 1 !
− x(− 2 cos(2x)) − 1 × (− 2 cos(2x)) dx
(x − 2) ln(3x) dx
!
!
= 12 x2 sin(2x) + 12 x cos(2x) − 12 cos(2x) dx
= (ln(3x))(x − 2) dx
= 12 x2 sin(2x) + 12 x cos(2x) − 12 × 12 sin(2x) + c !
1 2 1 1 2
1 2 1
= 4 (2x − 1) sin(2x) + 2 x cos(2x) + c. = ln(3x) × ( 2 x − 2x) − ( x − 2x) dx
x 2
!
1
Solution to Activity 37 = 2 x(x − 4) ln(3x) − ( 12 x − 2) dx
!
!
(5x + 2)e−5x dx 1 1
= 2 x(x − 4) ln(3x) − 2 (x − 4) dx
, - ! , -
1 1
= (5x + 2) e−5x − 5 e−5x dx = 12 x(x − 4) ln(3x) − 12 ( 12 x2 − 4x) + c
−5 −5
! = 12 x(x − 4) ln(3x) − 14 x(x − 8) + c.
−5x −5x
= − 15 (5x + 2)e + e dx
= − 15 (5x + 2)e−5x − 1 −5x

−5 e dx +c
= − 15 e−5x (5x + 2 + 1) + c
= − 15 (5x + 3)e−5x + c.
(An alternative way to find this integral is to start
by writing it as
! !
−5x
5 xe dx + 2 e−5x dx.
The first integral in this expression has an integrand
of the form xg(x), where g(x) is an expression that
you can integrate, so you can find this integral by
using integration by parts. The second integral is
straightforward to find.)
209
Solution to Activity 40 and

! hence
(a) We write e2x cos x dx = 15 e2x sin x + 25 e2x cos x + c.
! !
e sin(3x) dx = sin(3x) ex dx.
x
That is,
!
Integrating by parts twice gives e2x cos x dx = 15 e2x (sin x + 2 cos x) + c.
!
sin(3x) ex dx
! Solution to Activity 41
! 1 2 11 ! 1
= sin(3x)e − (3 cos(3x))ex dx
x
xe 3x 1 3x
dx = x × 3 e − 1 × 13 e3x dx
! 0 0 0
!
= sin(3x)e − 3 cos(3x)ex dx
x
1
) 3x '1 1 1 3x
= 3 xe 0 − 3 e dx
0
= sin(3x)ex 2 11
, ! -
x x
= 13 (e3 − 0) − 13 13 e3x
− 3 cos(3x)e − (−3 sin(3x))e dx 0
) '1
! = 13 e3 − 19 e3x 0
& +
= sin(3x)ex − 3 cos(3x)ex − 9 sin(3x)ex dx. = 13 e3 − 19 e3 − 1
So ! = 19 (3e3 − e3 + 1)
10 sin(3x) ex dx = 19 (2e3 + 1)
= 4.5746 (to 4 d.p.).
= sin(3x)ex − 3 cos(3x)ex + c
and! hence Solution to Activity 42
x (a) If x ∈ [π/12, π/6], then 3x ∈ [π/4, π/2].
sin(3x) e dx
Now
= 101
sin(3x)ex − 3
10 cos(3x)ex + c.
sin θ > 0 when θ ∈ [π/4, π/2],
That is,
! so
ex sin(3x) dx sin(3x) > 0 when x ∈ [π/12, π/6].
1 x 3 x Also
= 10 e sin(3x) − 10 e cos(3x) + c
= 1 x x > 0 when x ∈ [π/12, π/6].
10 e (sin(3x) − 3 cos(3x)) + c.
(b) Integrating by parts twice gives Hence
! x sin(3x) > 0 when x ∈ [π/12, π/6].
e2x cos x dx
That is, the graph of f (x) = x sin(3x) lies above
! the x-axis for values of x in the interval
= e sin x − (2e2x ) sin x dx
2x
[π/12, π/6].
!
(b) It follows from part (a) that the area between
= e2x sin x − 2 e2x sin x dx
this graph and the x-axis from x = π/12 to
= e2x x = π/6 is
, sin x ! - ! π/6
2x 2x
− 2 e (− cos x) − (2e )(− cos x) dx x sin(3x) dx.
! π/12
= e2x sin x + 2e2x cos x − 4 e2x cos x dx.

So !
5 e2x cos x dx = e2x sin x + 2e2x cos x + c
210
(c) Integrating by parts gives The function y = e−x is decreasing on its whole
! π/6 domain, so the function y = e−x − 1 is also
x sin(3x) dx decreasing on its whole domain. The x-intercept
π/12
! of its graph is given by e−x − 1 = 0, which gives
2 1π/6 π/6
1
= x(− 3 cos(3x)) − (− 13 cos(3x)) dx e−x = 1
π/12 π/12 − x = ln 1
! π/6
) 'π/6 −x=0
= − 13 x cos(3x) π/12 + 13 cos(3x) dx
π/12 x = 0.
. .π# . π ##
1 π π So its x-intercept is x = 0.
= −3 cos − cos
6 2 12 4
2 1π/6 So we have the following table of signs, for
+ 13 13 sin(3x) x > 0.
π/12
.π π 1 # ) ' x (0, 2) 2 (2, ∞)
π/6
= − 13 ×0− × √ + 19 sin(3x) π/12
6 12 2 x−2 − 0 +
π . .π# . π ## e−x − 1 − − −
= √ + 19 sin − sin
36 2 2 4 (x − 2)(e−x − 1) + 0 −
π . 1 #
= √ + 19 1 − √ The table shows that the graph of the function
36 2 2
= 0.0943 (to 4 d.p.). f (x) = (x − 2)(e−x − 1) lies above the x-axis for
0 < x < 2 and below the x-axis for x > 2.
(The area found in this solution is shown below.)
(b) Because the graph of f lies above the x-axis for
0 < x < 2 and below the x-axis for x > 2, to
find the total area between the graph of f and
the x-axis from x = 0 to x = 4 we have to work
out the signed area from x = 0 to x = 2 and the
signed area from x = 2 to x = 4 separately.
The signed area between the graph of f and the
x-axis from x = 0 to x = 2 is given by
! 2
(x − 2)(e−x − 1) dx
0
2 12 ! 2
−x
= (x − 2)(−e − x) − (−e−x − x) dx
0 0
! 2
= 0 − (−2)(−e−0 ) + (e−x + x) dx
0
2 12
−x 1 2
= 2(−1) + −e + 2 x
0
−2 −0
= −2 + (−e + 2) − (−e + 0)
Solution to Activity 43 = −2 − e−2 + 2 + 1
(a) To help us construct a table of signs for the = 1 − e−2 .
function f (x) = (x − 2)(e−x − 1), first we Since the graph of f lies above the x-axis for
consider the two factors of the expression 0 < x < 2, it follows that the area between the
(x − 2)(e−x − 1) separately. graph of f and the x-axis from x = 0 to x = 2 is
The function y = x − 2 is increasing on its 1 − e−2 .
whole domain, and the x-intercept of its graph
is x = 2.
211
! !
Similarly, the signed area between the graph (b) sin x cos x dx = 1
sin(2x) dx
2
of f and the x-axis from x = 2 to x = 4 is given
by = 1
× 12 (− cos(2x)) + c
! 4 2
(x − 2)(e−x − 1) dx = − 14 cos(2x) + c.
2
2 14 ! 4 Solution to Activity 45
= (x − 2)(−e−x − x) − (−e−x − x) dx du
2 2 Let u = sin x; then = cos x. So
! 4 ! dx
!
= 2(−e−4 − 4) − 0 + (e−x + x) dx sin x cos x dx = u du
2
2 14
= −2e −4
− 8 + −e−x + 12 x2 = 12 u2 + c
2
= −2e−4 − 8 + (−e−4 + 8) − (−e−2 + 2) = 12 sin2 x + c.
= −3e−4 + e−2 − 2. (There’s an alternative approach to finding this
integral, as follows. Let u = cos x; then
Since the graph of f lies below the x-axis for du
2 < x < 4, it follows that the area between the = − sin x. So
dx! !
graph of f and the x-axis from x = 2 to x = 4 is
−(−3e−4 + e−2 − 2) = 3e−4 − e−2 + 2. sin x cos x dx = cos x sin x dx
!
So the total area between the graph of f and = − (cos x)(− sin x) dx
the x-axis from x = 0 to x = 4 is !
(1 − e−2 ) + (3e−4 − e−2 + 2) = − u du
= 3 − 2e−2 + 3e−4
= − 12 u2 + c
= 2.784 (to 4 s.f.).
= − 12 cos2 x + c.
(The area found in this solution is shown below.)
This approach gives a different answer, but of
course it’s equivalent to the first answer. You can
confirm this by using the identity
sin2 θ + cos2 θ = 1, as follows:
− 12 cos2 x + c = − 12 (1 − sin2 x) + c
= 1
2 sin2 x − 1
2 +c
1 2
= 2 sin x + d,
where d = − 12 + c is an arbitrary constant.
Notice also that the solution to Activity 44(b) gives
a third
! different answer:
sin x cos x dx = − 14 cos(2x) + c.
Solution to Activity 44 You can confirm that this answer is equivalent to

! ! the first answer above by using the identity
(a) cos2 θ dθ = 1
2 (1 + cos(2θ)) dθ cos(2θ) = 1 − 2 sin2 θ, as follows:
! − 14 cos(2x) + c = − 14 (1 − 2 sin2 x) + c
1
= (1 + cos(2θ)) dθ
2 = 1
2 sin2 x − 1
4 +c
1 1 1 2
= 2 (θ + 2 sin(2θ)) + c = 2 sin x + d,
1
= 4 (2θ + sin(2θ)) + c. where d = − 14 + c is an arbitrary constant.)
212
(The diagram below shows the graphs of the three second expression is a simple function of a
different antiderivatives of f (x) = sin x cos x that linear expression.
!
were found in the solutions to Activity 44(b) and
Activity 45. The arbitrary constant has been taken (x + cos(5x)) dx = 12 x2 + 15 sin(5x) + c.
to be zero in each case. You can see that, as you’d
(d) You could use integration by substitution
expect, the three graphs appear to be vertical
(taking u = x2 + 3) or integration by parts, but
translations of each other.)
the simplest method is to start by multiplying
out! the brackets. !
x(x2 + 3) dx = (x3 + 3x) dx
= 14 x4 + 32 x2 + c
= 14 x2 (x2 + 6) + c.
1
(e) The integrand can be written as &√ +2 ,
1+ 3x
√
so it’s a function of the linear expression
√ 3 x.
So you can use the substitution u = 3 x,
1
together with the indefinite integral of ,
1 + x2
which is a standard integral.
√ √
Let u = 3 x; then du/dx = 3. So
! ! √
1 1 1
dx = √ &√ + × 3 dx
Solution to Activity 46 1 + 3x2 3 1+ 3x
2
!
(a) Since the derivative of 9x4 is 36x3 , which is 1 1
‘nearly’ x3 , you can use the substitution =√ du
3 1 + u2
u = 9x4 . 1
du = √ tan−1 u + c
Let u = 9x4 ; then = 36x3 . So 3
! dx ! 1 .√ #
= √ tan−1 3 x + c.
x3 cos(9x4 ) dx = 1
36 cos(9x4 ) × 36x3 dx 3
! (f) Since the derivative of 7 − x3 is −3x2 , which is
1
= 36 cos u du ‘nearly’ the numerator x2 , you can use the
1 substitution u = 7 − x3 .
= 36 sin u + c
du
= 1
sin(9x4 ) + c. Let u = 7 − x3 ; then = −3x2 . So
36
! dx !
(b) The integrand is of the form xg(x), where g is a x2 1
function that you can integrate, which suggests dx = − 1
3 (−3x2 ) dx
(7 − x3 )7 (7 − x3 )7
that you should try integration by parts. To !
1 1
integrate the ‘second’ expression, you can use = −3 du
u7
the fact that it is a simple function of a linear !
expression. = − 13 u−7 du
! !
1 1
x cos(5x) dx = 5 x sin(5x) − 5 sin(5x) dx 1 1 −6
=− × u +c
& + 3 −6
= 15 x sin(5x) − 15 − 15 cos(5x) + c 1
= +c
= 15 x sin(5x) + 1
25 cos(5x) + c. 18u6
(c) The integrand is the sum of two expressions 1
= + c.
that are straightforward to integrate. The 18(7 − x3 )6
213
(g) The integrand is of the form xg(x), where g is a du

Let u = 1 − 2ex ; then = −2ex . So
function that you can integrate, which suggests ! dx !
that you should try integration by parts. To
ex (1 − 2ex ) dx = − 12 (1 − 2ex )(−2ex ) dx
integrate the ‘second’ expression, you can use
!
the fact that it’s a simple function of a linear
expression. = − 12 u du
!
xe−x/3 dx = − 12 × 12 u2 + c
. # ! . # = − 14 u2 + c
−x/3
= x −3e − 1 × −3e−x/3 dx = − 14 (1 − 2ex )2 + c
!
−x/3 = − 14 (1 − 4ex + 4(ex )2 ) + c
= −3xe + 3 e−x/3 dx
= − 14 (1 − 4ex + 4e2x ) + c
= −3xe−x/3 − 9e−x/3 + c = ex − e2x − 1
+c
4
= −3e−x/3 (x + 3) + c. x x 2
= e − (e ) − 1
+c
4
(h) You can use the trigonometric identity x x
= e (1 − e ) − + c 1
4
sin(2θ) = 2 sin θ cos θ.
! ! = ex (1 − ex ) + d,
&1 + &1 + 1
sin 2 x cos 2 x dx = 2 sin x dx where d = − 14 + c is an arbitrary constant.)
! (k) You can start by multiplying out the brackets
= 12 sin x dx and applying the sum rule. Then you have two
integrals to find. One of them is
= − 12 cos x + c.
straightforward, and you can find the other by
(i) You can use the fact that the integrand is a using integration by substitution, taking u = x2 .
!
simple function of a linear expression, or you
can use the substitution u = x − 1. x(1 + sin(x2 )) dx
! !
(x − 1)5
(x − 1)4 dx = + c. = (x + x sin(x2 )) dx
5
! !
(j) You could use integration by substitution
(taking u = 1 − 2ex ), but the simplest method = x dx + x sin(x2 ) dx
is to start by multiplying out the brackets. !
! ! = 2 x + x sin(x2 ) dx
1 2
e (1 − 2e ) dx = (ex − 2(ex )2 ) dx
x x
!
Let u = x2 ; then du/dx = 2x. The expression
= (ex − 2e2x ) dx
above becomes
!
= ex − 2 × 12 e2x + c 1 2
(sin(x2 )) × 2x dx
1
2x + 2
= ex − e2x + c !
1 2 1
= ex − (ex )2 + c = 2 x + 2 sin u du
= ex (1 − ex ) + c. = 12 x2 + 12 (− cos u) + c
(Here’s how the integration goes if you use
= 12 x2 − 1
2 cos(x2 ) + c
integration by substitution.
= 12 (x2 − cos(x2 )) + c.
214
(l) Since the derivative of 3x2 − 2 is 6x, which is Solution to Activity 49

‘nearly’ x, you can use the substitution The area of the cross-section, in square centimetres,
u = 3x2 − 2. is
du ! 1.5 "
Let u = 3x2 − 2; then = 6x. So 1
x3 + 1 dx.
! dx! −1
3
2 8
x(3x − 2) dx = 1 2 8
(3x − 2) × 6x dx A computer gives the value of this definite integral
6
! as 0.939 (to 3 s.f.). So the volume of plastic
= 1
u8 du required to make one metre of the edging is
6
approximately 93.9 cm3 .
1 1 9
= 6 × 9u (Details of how to use the CAS to find the indefinite
= 1 2
2)9 + c. integral are given in the Computer algebra guide, in
54 (3x −
the section ‘Computer methods for CAS activities
Solution to Activity 48 in Books A–D’.)
(a) The domain of f is the interval
√ [−1, ∞) Solution to Activity 50
(because the expression x + 1 is defined only
when x is in this interval). The change in the displacement of the object from
time t = 0 to time t = 10 is
(b) The graph of f is shown below. ! 10 , 2 -
t
sin dt.
0 150
A computer gives the value of this definite integral
as 2.15 (to 2 d.p.).
So the displacement of the object at time t = 10 is
(6 + 2.15) m = 8.15 m, to the nearest centimetre.
(Details of how to use the CAS to find the indefinite
integral are given in the Computer algebra guide, in
the section ‘Computer methods for CAS activities
in Books A–D’.)
(c) The x-intercepts of f are 1.84 and 8.16 (to

3 s.f.).
√
(The exact values are 5 ± 10.)
(d) The area between the graph of f and the
x-axis, between the two x-intercepts, is
! 8.162... 2
x − 10x + 15
− √ dx.
1.837... 1+ x+1
A computer gives the value of this expression as
12.4 (to 3 s.f.).
(Details of how to use the CAS for this activity
are given in the Computer algebra guide, in the
section ‘Computer methods for CAS activities
in Books A–D’.)
215
Acknowledgements
Acknowledgements
Grateful acknowledgement is made to the following source:
Page 193: Mark Hobbs, for the idea for the cartoon
216

mst124 Unit8 E1i1 PDF

Uploaded by

Copyright:

Available Formats

mst124 Unit8 E1i1 PDF

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

mst124 Unit8 E1i1 PDF

Uploaded by

Copyright:

Available Formats

Unit 8

1 Areas, signed areas and deﬁnite

1.1 Areas and signed areas

Suppose that the architect wants to calculate the area shaded in

Figure 2 A collection of rectangles whose total area is approximately the

Example 1 Calculating an approximate value for an area

The calculation in Example 1 shows that a (rather crude) approximate

Activity 1 Calculating an approximate value for an area

Table 1 Approximate values for the area in Figure 1(b)

In general, if you have a continuous function f whose graph lies on or

Figure 5 Regions on a graph

Example 2 Understanding signed areas

Positive signed area

Negative signed area Solution

Activity 2 Understanding signed areas

Figure 6 (a) A collection of rectangles whose total signed area is

Figure 7 (a) Another instance of a collection of rectangles whose total

Activity 3 Calculating an approximate value for a signed area

In the next activity you can use a computer to explore approximations to

Activity 4 Calculating more approximate values for signed areas

So far in this subsection, whenever a signed area from x = a to x = b has

Figure 8 A signed area equal to zero

Suppose that f is a continuous function whose domain includes the

Figure 9 The graph of a continuous function f whose domain includes

Figure 10 Areas on a graph

Example 3 Understanding the extended deﬁnition of signed area

Activity 5 Understanding the extended deﬁnition of signed area

The graph in Example 3 is repeated below. In each of parts (a)–(g), use

1.2 Deﬁnite integrals

Figure 11 The graph of a function, and some areas

Standard properties of deﬁnite integrals

Figure 12 Two adjacent signed areas

Activity 6 Understanding deﬁnite integrals

Activity 7 Using a standard property of deﬁnite integrals

Consider again the graph in Activity 6.

Figure 13 Signed areas between the graphs of particular functions and

In fact, the input variable of the function in a deﬁnite integral is what’s

(Remember that the strategy on page 113 speciﬁes ‘endpoint of subinterval

Algebraic deﬁnition of a deﬁnite integral

If f is a particular continuous function, and a and b are numbers in its

Figure 15 (a) The subinterval method applied to a function f from

Joseph Fourier (1768–1830)

2 The fundamental theorem of

2.1 What is the fundamental theorem of calculus?

Figure 16 The graph of a continuous function f

Figure 17 A continuous function f and a signed area A(x)

For example, consider the constant function f (x) = 3, whose graph is

Figure 19 The graph of the constant function f (x) = 3

Figure 20 shows some values of A(x).

Figure 20 Some values of A(x) when f is the constant function f (x) = 3

Figure 21 Signed-area-so-far functions for constant functions, with

Figure 22 A signed-area-so-far function for a function f that isn’t a

Figure 23 (a) When x is close to 5, the rate of change of A(x) is about 3

Figure 24 The instantaneous rate of change of A(x) at x is f (x)

Now let F be any antiderivative of f . Then, since A and F are both

Fundamental theorem of calculus

This theorem can be stated in words as follows.

Fundamental theorem of calculus (in words)

The equation in the fundamental theorem of calculus is true no matter