6.2 integration by substitution m.l.king jr. birthplace, atlanta, ga greg kelly hanford high school...
Post on 27-Mar-2015
215 Views
Preview:
TRANSCRIPT
6.2Integration bySubstitution
M.L.King Jr. Birthplace, Atlanta, GA Greg KellyHanford High SchoolRichland, WashingtonPhoto by Vickie Kelly, 2002
The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.
Example 1:
52x dx Let 2u x
du dx5u du61
6u C
62
6
xC
The variable of integration must match the variable in the expression.
Don’t forget to substitute the value for u back into the problem!
Example:(Exploration 1 in the book)
21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integrand.
The derivative of is .21 x 2 x dx
1
2 u du3
22
3u C
3
2 22
13
x C
2Let 1u x
2 du x dx
Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.
Example 2:
4 1 x dx Let 4 1u x
4 du dx
1
4du dx
Solve for dx.1
21
4
u du3
22 1
3 4u C
3
21
6u C
3
21
4 16
x C
Example 3:
cos 7 5 x dx7 du dx
1
7du dx
1cos
7u du
1sin
7u C
1sin 7 5
7x C
Let 7 5u x
Example: (Not in book)
2 3sin x x dx 3Let u x23 du x dx
21
3du x dx
We solve for because we can find it in the integrand.
2 x dx
1sin
3u du
1cos
3u C
31cos
3x C
Example 7:
4sin cos x x dx
Let sinu x
cos du x dx
4sin cos x x dx
4 u du51
5u C
51sin
5x C
Example 8:
24
0tan sec x x dx
The technique is a little different for definite integrals.
Let tanu x2sec du x dx
0 tan 0 0u
tan 14 4
u
1
0 u du
We can find new limits, and then we don’t have to substitute back.
new limit
new limit
12
0
1
2u
1
2We could have substituted back and used the original limits.
Example 8:
24
0tan sec x x dx
Let tanu x
2sec du x dx4
0 u du
Wrong!The limits don’t match!
42
0
1tan
2x
2
21 1tan tan 0
2 4 2
2 21 11 0
2 2
u du21
2u
1
2
Using the original limits:
Leave the limits out until you substitute back.
This is usually more work than finding new limits
Example: (Exploration 2 in the book)
1 2 3
13 x 1 x dx
3Let 1u x
23 du x dx 1 0u
1 2u 1
22
0 u du
23
2
0
2
3u Don’t forget to use the new limits.
3
22
23
22 2
3 4 2
3
Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators.
In another generation or so, we might be able to use the calculator to find all integrals.
You must practice finding integrals by hand until you are good at it!
top related