Definite Integral
Definite Integral
Definite Integral
INTEGRAL
The Definite Integral
Objectives
b
2. Interpret the expression ∫a ƒ(x) dx when ƒ(x) ≥ 0 and a ≤ b.
3. Relate the definite and indefinite integrals through the
fundamental theorem of calculus.
The Definite Integral
Objectives
4. Evaluate the definite integral of a function applying the
rules of integration.
Example 1.
Find the area of a region bounded by the function ƒ(x)=3 (a
constant function), the x-axis, the vertical line x=0 (the y-axis),
and x=4.
Solution:
We first locate the region in the
xy-plane. The graph of the
function ƒ is a line parallel to
the x-axis. Since ƒ(0)=3 and
ƒ(4)=3, then the points (0,3) and A
(4,3) are on the line. Note that
a=0 and b=4. The shaded region
is in the form of a rectangle
whose length l is 3 and width w
is 4.
Regions Bounded by Polygons
Example 1.
Find the area of a region bounded by the function ƒ(x)=3 (a
constant function), the x-axis, the vertical line x=0 (the y-axis),
and x=4.
Solution:
A = lw
= 3(4) A
= 12
Regions Bounded by Polygons
Example 2.
Solve for the area of a region bounded from above by the
function ƒ(x)=x + 3, the x-axis, the y-axis (x=0) and the vertical
line x=4.
Solution:
The graph of the function
ƒ(x)=x + 3 is a line whose slope
is 1 and y-intercept is 3, so it
passes through point (0,3).
Since ƒ(4)=4+3=7, then the line
also passes through (4,7). In
this region, a=0 and b=4.
Regions Bounded by Polygons
Example 2.
Solve for the area of a region bounded from above by the
function ƒ(x)=x + 3, the x-axis, the y-axis (x=0) and the vertical
line x=4.
Solution:
The region is in the form of a
trapezoid whose parallel sides
(b1 and b2) have lengths 3 and 7.
The distance h between these
parallel sides is 4. Thus, the
area is
A = (b1 + b2)h = (3 + 7)(4)
2 2
A = 20
Approximating the
Area Under the Curve
Approximating the Area Under the Curve
Sn = A1 + A2 + A3 + … + An
or
Example 3.
Calculate an approximation of the area of the region An
bounded by the function ƒ(x)=x2, the x-axis, and the vertical
lines x=1 and x=3. Approximate the areas using 4 subintervals.
Solution:
The graph of the function
ƒ(x)=x2 is a parabola. The
vertical lines are x=1 and x=3,
thus a=1 and b=3. Since ƒ(1)=1
and ƒ(3)=9, then the graph is a
curve that crosses the points
(1,1) and (3,9). The shaded
region is shown in the figure.
Approximating the Area Under the Curve
In the figure, we
note that ƒ(ci) may
be positive or
negative. When ƒ(ci)
is positive, the
product ƒ(ci) x is
the area of the
rectangle Ri. When
ƒ(ci) is negative, the
product ƒ(ci) x is
the negative of the
area of the rectangle.
If we take the sum
of these products, Sn = ƒ(c1) x + ƒ(c2) x + … + ƒ(cn) x
we obtain
The Definite Integral
∫a ƒ(x)dx =nlim S
b
∞
n
∫a ∫a
b b
ƒ(u)du or ƒ(t)dt
Area Under the Curve
Area Under the Curve
A = ∫a ƒ(x)dx
b
Area Under the Curve
Example 4.
Write as a definite integral the area of the region bounded by
the curve ƒ(x)=5x, the x-axis, and the vertical lines x=3 and x=4.
Solution:
The integrand is the
function ƒ(x)=5x. The
limits of integration are
a=3 and b=4.
Therefore, as a definite
integral, the area A is
denoted as
A = ∫3 5(x)dx
4
Area Under the Curve
We see that the boundaries are still the same: the x-axis and
the vertical lines x=a and x=b. Only this time, the function ƒ is
negative in the closed interval [a, b]. To solve for the area of the
shaded region, we obtain the negative of the definite integral of
the function from a to b. That is, when ƒ(x) is negative in [a, b],
then the area A is
A = - ∫a ƒ(x)dx
b
1. ZERO INTEGRAL
When the limits of integration are equal (b=a), then the
value of the integral is equal to zero. That is,
b
∫a ƒ(x)dx = 0
Example: ∫1 x dx = 0 because the limits of integration are
1
2
equal.
Properties of the Definite Integral
2. ORDER OF INTEGRATION
We define the integral from b to a as the negative of the
integral from a to b. That is,
∫b ƒ(x)dx = - ∫a ƒ(x)dx
a b
4. SUM INTEGRAL
The definite integral of the sum of two functions ƒ and g is
the sum of the integrals of the two functions.
∫a [ƒ(x)+g(x)]dx = ∫a ƒ(x)dx + ∫a
b b b
g(x)dx
5. DIFFERENCE INTEGRAL
The integral of the difference of two functions ƒ and g is the
difference of the integrals of the two functions.
∫a [ƒ(x)-g(x)]dx = ∫a ƒ(x)dx - ∫a g(x)dx
b b b
Example: ∫ (x -x )dx = ∫ x dx - ∫ x dx
3 3 3
3 2 3 2
1 1 1
Properties of the Definite Integral
7. ADDITIVE INTEGRAL
If a < c < b, then
b c
∫a ƒ(x)dx = ∫a ƒ(x)dx + ∫c ƒ(x)dx b
Example: ∫15
x2dx = ∫1
3
x2dx + ∫3 x dx
5
2
Fundamental Theorem
of Calculus
Fundamental Theorem of Calculus
∫a
b
ƒ(x)dx = F(b) – F(a)
This means that the definite integral from x=a to x=b is equal
to an antiderivative of ƒ evaluated at x=b minus this
antiderivative of ƒ evaluated at x=a. The difference F(b)-F(a) is
often denoted by the shorthand expression
b
F(b) – F(a) = F(x)|a
Fundamental Theorem of Calculus
1. Obtain an antiderivative of F of ƒ.
Example 5.
Evaluate ∫2 3xdx.
4
Solution:
The integrand is ƒ(x)=3x.
The limits of integration are a=2 and b=4.
4
Therefore, ∫2 3xdx = 18.
Fundamental Theorem of Calculus
∫ 3xdx
4
A= 0
= 3 ∫ dx
4
0
4
= 3x |0
= 3(4-0)
= 12
Applications of the Definite Integral
Example 7.
We find the area of the region
bounded by ƒ(x)=x2, the vertical
lines x=1 and x=3. The integrand is
ƒ(x)=x2 and the limits of
integration are a=1 and b=3. Thus,
the area of the shaded region is
A= ∫ x dx
1
3
2
3
= x3 /3 |1
= 33 - 13
3 3
= 27/3 - 1/3
= 26/3 _
= 8.666666
Velocity and
Acceleration
Velocity and Acceleration
Example 9.
Starting at rest, the acceleration of a body moving in rectilinear
motion is given by the acceleration function
a(t) = 10t – 3 m/s2
a. Find the velocity function of the body.
b. What is the distance covered between 3 seconds and 5
seconds?
Solution:
a. We obtain the velocity function of the body by getting the
integral of the acceleration function.
=[ 5(5)3 – 3(5)2
3 2 ]- [ 5(3)3 – 3(3)2
3 2 ]
= 139.3 m
Important Terms
Summary:
The definite integral is the area of the region bounded
by the function ƒ(x), the x-axis, and the vertical lines
x=a and x=b. In symbols, the definite integral is denoted,
∫ab
ƒ(x)dx
In evaluating the definite integral, the fundamental
theorem of calculus is applied where the definite
integral of a function ƒ(x) over the interval of x [a, b] is
the antiderivative F(x) of the function evaluated at b
minus the antiderivative F(x) at a. In symbols,
∫a ƒ(x)dx = F(x)| a
b b
= F(b) – F(a)
Summary
Summary:
The definite integral is the algebraic sum of areas
where regions above the x-axis are considered positive
and regions below the x-axis negative.
The following rules of integration apply to the definite
integral:
a. Constant multiplier rule
b. Sum rule
c. Difference rule
Summary
Summary:
Given the rate of change of a quantity, the definite
integral allows the computation of interval values of the
quantity.
For example, given the velocity as a function of time, we
can solve for the position of a moving body for certain
time intervals applying the definite integral.