Chapter 6
Integration
Section 4
Learning Objectives for Section 6.4 The
Definite Integral
1. The student will be able to
approximate areas by using left and right sums.
2. The student will be able to
compute the definite integral as a limit of sums.
3. The student will be able to apply the properties of the definite
Introduction
We have been studying the indefinite integral or
antiderivative of a function.
∫
abf
(
x
)
dx
We now introduce the
definite integral. This integral will be the area
bounded by f (x), the x axis, and the vertical lines x = a
Estimating
One way to approximate the area under a curve is by filling the region with
rectangles and calculating the sum of the areas of the rectangles.
Take the width of each rectangle to be
x = 1. If we use the left endpoints, the heights of the four rectangles are
f (1), f (2), f (3) and f (4), respectively.
L4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx
= 2.5 + 4 + 6.5 + 10 = 23
f (x) = 0.5 x2 + 2
1 2 3 4 5
dx
x
2
5
.
0
5
1
2
+
Estimating Area
(continued)
We can repeat this using the right side of each rectangle to determine the height.
The width of each rectangle is again
x = 1. The heights of each of the four rectangles are now f (2), f (3), f (4) and
f (5), respectively.
The sum of the rectangles is then
R4 = 4 + 6.5 + 10 + 14.5 = 35
f (x) = 0.5 x2 + 2
1 2 3 4 5
Estimating Area
(continued)
The previous average of 29 is very close to the actual area of 28.666….
Our accuracy can be improved if we increase the number or rectangles, and let x get smaller.
f (x) = 0.5 x2 + 2
1 2 3 4 5
The error in our process can be calculated if the function is
monotone. That is, if the function is only increasing or only decreasing.
Estimating Area
(continued)
If the function is increasing, convince yourself by looking at the picture that
Ln Area Rn
If f is decreasing, the inequalities go the other way.
f (x) = 0.5 x2 + 2
1 2 3 4 5
If you use Ln to estimate the area, then
Error = |Area – Ln| |Rn – Ln|. If you use Rn to estimate the area, then
Theorem 1
It is not hard to show that
|Rn – Ln| = | f (b) – f (a)| x,
and that for n equal subintervals,
For our previous example: 12
4 1 5
| 5 . 2 5
. 14 |
Error ≤ − − =
n
a
b
a
f
b
f
−
⋅
−
≤
|
(
)
(
)
|
Error
Theorem 1
.
n a b x = −
Theorem 2
If f (x) is either increasing or decreasing on [a, b], then its left and right sums approach the same real number I as n .
This number I is the area between the graph of f and the x axis from
Approximating Area
In the previous example, we had f (x) = 0.5 x 2 + 2
1 2 3 4 5
12
4
1
5
|
5
.
2
5
.
14
|
Error
≤
−
−
=
If we wanted a particular accuracy, say 0.05, we could use the error
formula to calculate n, the number of rectangles needed:
05
.
0
1
5
|
5
.
2
5
.
14
|
−
−
=
n
Definite Integral as Limit of Sums
We now come to a general definition of the definite integral.
Let f be a function on interval [a, b]. Partition [a, b] into n
subintervals at points
a = x0 < x1 < x2 < … < xn–1 < xn = b.
The width of the kth subinterval is x
k = (xk – xk-1). In each subinterval, choose an arbitrary point ck
Definite Integral as Limit of Sums
(continued)
∑
= −
− Δ = Δ
+ + Δ + Δ = n k k n
n f x x f x x f x x f x x
L
1
1 1
1
0) ( ) . . . ( ) ( )
(
∑
=Δ
=
Δ
+
+
Δ
+
Δ
=
n k k nn
f
x
x
f
x
x
f
x
x
f
x
x
R
1 2
1
)
(
)
.
.
.
(
)
(
)
(
Then defineSn is called a Riemann sum. Notice that Ln and Rn are both special cases of a Riemann sum.
∑
=Δ
=
Δ
+
+
Δ
+
Δ
=
n k k nn
f
c
x
f
c
x
f
c
x
f
c
x
S
1 2
1
)
(
)
.
.
.
(
)
(
)
A Visual Presentation
of a Riemann Sum
a = x 0 x 1 x 2 . . . x n – 1 x n = b
∑
=
Δ
n
k
k k
x
c
f
1
)
(
c 1 c 2 c n f (c 1)
f (c 2) Δx
The area under the curve is
Area (Revisited)
Let’s revisit our original problem and calculate the Riemann sum using the midpoints for ck.
The width of each rectangle is again
x = 1. The heights of the four
rectangles are now f (1.5), f (2.5),
f (3.5) and f (4.5), respectively. The sum of the rectangles is then
S4 = 3.125 + 5.125 + 8.125 + 12.125 = 28.5
f (x) = 0.5 x2 + 2
1 2 3 4 5
The Definite Integral
Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as max xk 0.
This limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted
The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b.
dx
x
f
b
a
Negative Values
If f (x) is positive for some values of x
on [a, b] and negative for others, then the definite integral symbol
represents the cumulative sum of the signed areas between the graph of
f (x) and the x axis, where areas above are positive and areas below negative.
∫
abf
(
x
)
dx
B
A
dx
x
f
b
a
=
−
+
∫
(
)
y = f (x)
a
b A
Examples
Calculate the definite integrals by referring to the figure with the indicated areas.
Area A = 3.5
Area B = 12
5
.
3
)
(
=
−
∫
abf
x
dx
5
.
8
12
5
.
3
)
(
=
−
+
=
∫
acf
x
dx
y = f (x)
a
b A
B
c
12
)
(
=
Definite Integral Properties
[
f
(
x
)
±
g
(
x
)]
dx
=
a b
∫
af
(
x
)
dx
b
∫
±
ag
(
x
)
dx
b∫
f
(
x
)
dx
=
a b
∫
af
(
x
)
dx
c∫
+
cf
(
x
)
dx
b∫
k
⋅
f
(
x
)
dx
=
a b
∫
k
⋅
f
(
x
)
dx
a b
∫
f
(
x
)
dx
=
a b
∫
−
f
(
x
)
dx
b a
∫
f
(
x
)
dx
= 0
a a
Examples
Assume we know that
,
2
9
3
0
=
∫
x
dx
9
,
and
30
2
=
∫
x
dx
.
3
37
4 3
2
=
∫
x
dx
A)
4
4
34
(
9
)
36
0
2 3
0
2
=
=
=
∫
∫
x
dx
x
dx
B)
∫
−
=
∫
−
∫
3=
0 3 0 2 3 0
2
2
)
3
2
3
(
x
x
dx
x
dx
x
dx
18
)
2
9
(
2
)
9
(
3
−
=
Examples
(continued)
,
2
9
3 0=
∫
x
dx
9
,
and
30
2
=
∫
x
dx
3437
3
.
2=
∫
x
dx
C)
3
37
4 3 2 3 4 2−
=
−
=
∫
∫
x
dx
x
dx
D)
0
4 4
2
=
∫
x
dx
E)
64
3
37
3
9
3
3
3
3
4 3 2 3 0 2 4 0 2=
⋅
+
⋅
=
+
=
∫
∫
Summary
■
We summed rectangles under a curve using both the left and right ends and the centers and found that as the number of rectangles increased, accuracy of the area under thecurve increased.
■
We found error bounds for these sums.■
We defined the definite integral as the limit of these sums and found that it represented the area between the function and the x axis.