Chapter 7
Additional
Integration
Topics
Section 4
Integration Using
Tables
Learning Objectives for Section 7.4
Integration Using Tables
The student will be able to
■ Use a table of integrals.
■ Use substitution and
reduction formulas.
■ Solve application
problems.
2
Using a Table of Integrals
Table II of
Appendix C
contains integral
formulas
illustrating some
basic integrations.
More extensive
tables are available
for other integrals.
3
Example
x
1
dx
16  x
u
Table II, formula 27 fits:
a  bu  a
1
1
du 
ln |
|
a  bu
a
a  bu  a
with u = x, du = dx, a = 16 and b = 1.
x
16  x  4
1
1
dx  ln |
| C
4
16  x
16  x  4
4
Substitution and Integral Tables
Sometimes the formula matches exactly, as in the
preceding example.
Sometimes a substitution needs to be made in order to
fit one of the formulas on the table.
5
Example
2
2
x
9
x
 1 dx

This almost fits formula 41:
u
2
1
u  a du 
8
2
2


 u 2u 2  a 2 u 2  a 2  a 4 ln u  u 2  a 2 


If u = 3x , u2 = 9x2, du = 3dx and a = 1, we could make the
necessary adjustments.
6
Example
(continued)
x
1  1
9 x  1 dx     9 x 2 9 x 2  1 3dx
9  3
2
2
1
2
2

u
u
 1du

27

1 1

3x 18x 2  1 2
27 8 

9x 2  1 2  ln 3x  9x 2  1 2   C

7
Reduction Formulas
Sometimes using the table will not solve the integral
directly, but instead replaces the given integral with one
that has an exponent reduced by 1.
This type of formula is called a reduction formula and
means we need to apply the formula in the table
repeatedly until the integral is completely evaluated.
8
Example

x
Formula 47 fits:
n
au
u
e
n
n
a
u
x e dx  u e du 
  u n  1 e a u du
a
a
3 x
x e
3
3 x
2
x

x
e
dx
First use:  x e dx 

1
1
3
3 x
3 x
2
x
x
x
e
dx


x
e
3
x
e

6
x
e
dx
Second use: 

Third use:
3 x
3 x
2 x
x
e
dx


x
e
3
x
e 

 6 x e  x 6e  x  C
This last integration was done by parts!
9
Application: Producers’ Surplus
Find the producers’ surplus at a price level of $20 for the pricesupply equation p  S ( x)  5 x
Step 1. Find
x
500  x
, the supply when the price is $20
p  20 :
5x
p
500  x
5x
20 
500  x
10,000  20x  5x
x  400
10
Application
(continued)
p  S ( x) 
Step 2. Sketch a graph:
Step 3. Find the producers’
surplus (the shaded area in the
x
graph.
p  20
PS   [ p  S(x)]dx
5x
500  x
400
x
0
400


0
400


0

5x 
 20  500  x  dx
10,000  25x
dx
500  x
11
Application
(continued)
Use formula 20 with a = 10,000, b = –25, c = 500, and d = –1:
a  bu
bu ad  bc
 c  du du  d  d 2 ln c  du
PS  25x  2,500ln 500  x |0400


 10,000  2,500ln 100  2,500ln 500
 $5,976
12
Summary
■ There are tables in the appendix that contain formulas to
assist us in integration.
■ Sometimes we need to make substitutions before we use
these tables.
■ Sometimes these formulas need to be applied repeatedly in
order to complete an integration.
■ Along with previously learned methods of integration we
now have a much better repertoire for integrating more
complicated functions.
13