Revision Test 2

MAT1A3E: Revision
Semester Test 2
Breakdown:
Chapter 3.11 and 4.4: 16 Marks
Definition of the Hyperbolic Functions:
ex − e−x 1
sinh(x) = csch(x) =
2 sinh x
ex + e−x 1
cosh(x) = sech(x) =
2 cosh x
sinh x cosh x
tanh(x) = coth(x) =
cosh x sinh x
ran(sinh(x)) = (−∞, ∞)
ran(cosh(x)) = [1, ∞)
Hyperbolic Identities:
sinh(−x) = − sinh(x) cosh(−x) = cosh(x)
cosh2 (x) − sinh2 (x) = 1
1 − tanh2 (x) = sech2 (x)

sinh(x + y) = sinh x cosh y + cosh x sinh y
cosh(x + y) = cosh x cosh y + sinh x sinh y
The proofs of all the identities follow directly from
the definition of cosh(x) and sinh(x).
Calculating Limits Using the Limit Laws
Example 1
Use the Limit Laws and the graphs of f and g in Figure 1 to
evaluate the following limits, if they exist.
a. lim [f (x) + 5g(x)]

x→−2
b. lim [f (x)g(x)]
x→1
f (x)
c. lim
x→2 g(x)
Other important limit laws
Other important limit laws
Definition of a limit at infinity
Definition of a limit at infinity
Indeterminate forms
L’Hospital’s Rule:
Suppose f and g are differentiable and g 0 (x) 6= 0 when x is
near a. Suppose that
lim f (x) = 0 and lim g(x) = 0

x→a x→a
OR
lim f (x) = ±∞ and lim g(x) = ±∞.
x→a x→a
f (x) f 0 (x)
Then lim = lim 0
x→a g(x) x→a g (x)
if the RHS exists (or is ∞ or −∞).

Note: L’Hospital’s Rule also applies to the following:
x → a+ x → a−
x→∞ x → −∞
That is, it can also be used for one-sided limits and for limits at
infinity.
What if we want to find
lim [f (x).g(x)]
x→a
The method we use here is to write the product as a quotient.

That is: f g
f ·g = or f · g =
1/g 1/f
0 ∞
Then we have an indeterminate form of type or .
0 ∞
Note: you have to make a choice when converting f · g into a
fraction. The first option you should try is to leave the function
with the more complicated reciprocal in the numerator.
Indeterminate powers
I 00
I ∞0
I 1∞
To solve these, we use the natual logarithm!
y = f (x)g(x) =⇒ ln(y) = g(x) ln(f (x))
NB: we calculate lim ln(y), but this is not our final answer! We
x→a
are interested in:
lim y = lim eln(y) = elimx→a ln(y)

x→a x→a
Useful Functions
Antiderivatives:
A function F is an antiderivative of f on an interval I if
F 0 (x) = f (x)
for all x ∈ I.
If F is an antiderivative of f on an interval I, then the most

general antiderivative of f on I is F (x) + C where C is an
arbitrary constant.
Function Particular anti-derivative
cf (x) cF (x)
f (x) + g(x) F (x) + G(x)
1
xn (n 6= −1) n+1 x
n+1
1
`n|x|
x
ex ex
Function Particular antiderivative
cos x sin x
sin x − cos x
sec2 x tan x
sec x tan x sec x

Function Particular antiderivative
1
√ arcsin x
1 − x2
1
arctan x
1 + x2
cosh x sinh x
sinh x cosh x
In order to find a general way of finding
antiderivatives, we first have to understand the
notation used when we define them.
Areas types:
Left endpoint or Ln :
x1 = a + 1∆x,
x2 = a + 2∆x,
...,
xn = a + n∆x
xi = a + i∆x
n
b−a X
∆x = Ln = ∆xf (xi−1 )
n i=1
Right endpoint or Rn :
x1 = a + 1∆x,
x2 = a + 2∆x,
...,
xn = a + (n)∆x
xi = a + i∆x
n
X
b−a Rn = ∆xf (xi )
∆x =
n i=1
Midpoint or Mn :
1
x1 = a + ∆x,
2
3
x2 = a + ∆x,
2
...,

1
xn = a + (n − 1) + ∆x
2

1
=a+ n− ∆x
2

2i + 1
xi = a + ∆x
b−a 2
∆x =
n
n
X
Mn = ∆xf (xi )
i=0
Summation Rules:
What happens when n → ∞?
lim Rn = lim Ln = lim Mn

n→∞ n→∞ n→∞
b−a
Let f be defined for a 6 x 6 b and let ∆x = .
n
Let x∗i be a sample point in the i-th subinterval.
The definite integral of f from a to b is
Z b Xn
∗
f (x) dx = lim f (xi ).∆x
a n→∞
i=1
The definite integral of f from a to b exists

provided that the limit on the RHS exists. If the
limit (and hence the integral) exists, we say that f
is integrable on [a, b].
Properties of the definite integral
Rb
If we think of a f (x) dx purely as the limit-based
definition then we get:
Z a Z b
f (x) dx = − f (x) dx
b a
and Z a
f (x) dx = 0
a
Properties of the Integral
We assume f and g are continuous. Then
Z b
1. c dx = c(b − a) for c any constant
a
Z b Z b Z b

2. f (x) + g(x) dx = f (x) dx + g(x) dx
a a a
Z b Z b
3. cf (x) dx = c f (x) dx for c any constant
a a
Z b Z b Z b

4. f (x) − g(x) dx = f (x) dx − g(x) dx
a a a
Z c Z b Z b
5. f (x) dx + f (x) dx = f (x) dx
a c a
Comparison Properties of the Integral
Z b
6. If f (x) > 0 for a 6 x 6 b, then f (x) dx > 0
a
7. If f (x) > g(x) for a 6 x 6 b, then
Z b Z b
f (x) dx > g(x) dx
a a
8. If m 6 f (x) 6 M for a 6 x 6 b, then

Z b
m(b − a) 6 f (x) dx 6 M (b − a)
a
Fundamental Theorem of Calculus
Part 1: If f is continuous on [a, b], then the

function g defined by
Z x
g(x) = f (t) dt a6x6b
a
is continuous on [a, b] and differentiable on (a, b)

and g 0 (x) = f (x).
d x
Z
That is: f (t) dt = f (x)
dx a
Fundamental Theorem of Calculus
Part 2: If f is continuous on [a, b], then
Z b
f (x) dx = F (b) − F (a)
a
where F is any antiderivative of f , that is

F0 = f.
True or False?
1. If lim f (x) = 1 and lim g(x) = ∞ then
x→∞ x→∞
lim [f (x)]g(x) = 1.
x→∞
2. If F (x) is an antiderivative of f (x), then f 0 (x) = F (x).

3. If a function is concave down (for example looks like
y = −x2 + 12), we can obtain an overestimate of area by
applying the right endpoint rule.
4. If the functions f (x) and g(x) have the same
antiderivative then f (x) = g(x).
R3
5. The value of the integral 0 (x − 1) dx = 1. This gives
that the area under the y = x − 1 from 0 to 3 is 1.
Some Practice Problems:
1. If f ” (x) = x3 + sinh x with f (0) = 1 and f (2) = 2, find f (x).

2. Evaluate the limit:
arctan x − x
lim
x→0+ x arctan x
.
3. Estimate the area under the curve y = −3x2 + 2x − 1 on the

interval [−4, 0] using four rectangles and left end points.
Simplify your answer fully.
R0
4. Evaluate 3 3x2 dx by interpreting the integral as a limit of
sums.
A slightly more interesting one:
Rx 2
The function defined by erf(x) = √2π 0 e−t dt is called the error
function.
Ra Ra √
1 π
1. Show that a cos2 x sin2 x + b dt = 2 [erf(a) − erf(b)]
et2
2. Find the derivative
R a of the function
g(x) = erf(x) + a cos x sin2 x
2

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MAT1A3E: Revision

sinh(−x) = − sinh(x) cosh(−x) = cosh(x)

cosh2 (x) − sinh2 (x) = 1

1 − tanh2 (x) = sech2 (x)

a. lim [f (x) + 5g(x)]

lim f (x) = 0 and lim g(x) = 0

if the RHS exists (or is ∞ or −∞).

The method we use here is to write the product as a quotient.

y = f (x)g(x) =⇒ ln(y) = g(x) ln(f (x))

lim y = lim eln(y) = elimx→a ln(y)

If F is an antiderivative of f on an interval I, then the most

f (x) + g(x) F (x) + G(x)

sec x tan x sec x

lim Rn = lim Ln = lim Mn

The definite integral of f from a to b exists

8. If m 6 f (x) 6 M for a 6 x 6 b, then

Part 1: If f is continuous on [a, b], then the

is continuous on [a, b] and differentiable on (a, b)

where F is any antiderivative of f , that is

2. If F (x) is an antiderivative of f (x), then f 0 (x) = F (x).

1. If f ” (x) = x3 + sinh x with f (0) = 1 and f (2) = 2, find f (x).

3. Estimate the area under the curve y = −3x2 + 2x − 1 on the

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