chapter 7 additional integration topics section 4 integration using tables
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Chapter 7
Additional Integration Topics
Section 4
Integration Using Tables
2
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.
3
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.
4
Example
Table II, formula 27 fits:
with u = x, du = dx, a = 16 and b = 1.
dxxx 16
1
||ln11
abua
abua
adu
ubau
Cx
xdx
xx
|
416
416|ln
4
1
16
1
5
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.
6
Example
This almost fits formula 41:
If u = 3x , u2 = 9x2, du = 3dx and a = 1, we could make the necessary adjustments.
dxxx 19 22
2242222222 ln2
8
1auuaauauuduauu
7
Example(continued)
1
27u2 u2 1du
1
27
1
83x 18x 2 12 9x2 12 ln 3x 9x 2 12
C
x 2 9 x2 1 dx
1
9
1
3
9 x 2 9 x2 1 3dx
8
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.
9
Example
Formula 47 fits:
This last integration was done by parts!
dxex x3
First use:
Second use:
Third use:
un e au du
un e au
a
n
aun 1 e au du
x3 e x dx
x3 e x
1
3
1x2 e x dx
x3 e x dx x3 e x 3 x2 e x 6 x e x dx
6 x e x 6e x C
x3 e x dx x3 e x 3 x2 e x
10
Application: Producers’ Surplus
Find the producers’ surplus at a price level of $20 for the price-supply equation
Step 1. Find , the supply when the price is $20
5( )
500
xp S x
x
x
p 20 :
p 5x
500 x
20 5x
500 x
10,000 20x 5x
x 400
11
Application(continued)
Step 2. Sketch a graph:
Step 3. Find the producers’ surplus (the shaded area in the graph.
20p
400 x
5( )
500
xp S x
x
PS [ p S(x)]dx0
x
20 5x
500 x
dx0
400
10,000 25x
500 xdx
0
400
12
Application(continued)
Use formula 20 with a = 10,000, b = –25, c = 500, and d = –1:
a bu
c du du bu
d
ad bc
d 2ln c du
PS 25x 2,500ln 500 x |0400
10,000 2,500ln 100 2,500ln 500
$5,976
13
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.
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