Calc 1
Mon 12/1/08
6.8 - De▯nite Integrals with Substitution
Recall from 6.3 (substitution) that if we had an inde▯nite integral of the form
Z
f(g(x))g (x) dx
then we could rewrite that guy in terms of u, namely
Z
f(u) du
for u = g(x).
Now let’s see what happens when we slap some end points on there. ie,
Z
b 0
f(g(x))g (x) dx
a
There are two ways to go about doing this. You can do the u-substitution, evaluate the integral,
get everything back in terms of x and then plug in the end points. This would be like solving for
the inde▯nite integral (remember, always put the inde▯nite integrals back in terms of x (or
whatever variable you started with)!) and then just plug in b and a and subtract in the end.
There is nothing wrong with this method. It just might be an extra step or two, that’s all.
The other method (don’t worry, i’ll be doing an example soon if you’re confused!) involves keeping
u in there, but changing the end points. In other words, you have
Z b
f(u) du
a
However, keep in mind that a and b correspond to x, not u. A lot of times it’s helpful to write it
like this: Z
x=b
f(u) du
x=a
so that you know that the a and b are associated with x. How do you change the limits? Just
throw them in your g(x) (ie, u)! Thus,
Z g(b)
f(u) du
g(a)
This second method is nice because you don’t have to change everything back in terms of x
(because instead you change the limits). As I said before, either method works ▯ne. I am going to
try and solve every problem today using both methods and whichever one you like better you can
choose. As for the homework, the ▯rst bunch of problems ask you to use both methods, so you’ll
have to learn how to do both methods. On an exam, though, you’ll just be asked to evaluate the
integral and can use any (valid) method you want.
1 Example: Evaluate Z
1 p
2x 1 + x
0
u = 1 + x , so du = 2x dx.
Method 1: Z
x=1 p
= u du
x=0
x=1
= 2u3=2▯
3 x=0
▯
2 2 3=2▯
= 3(1 + x ) 0
2 2 3=2 2 2 3=2
= (1 + 1 ) ▯ (1 + 0 )
3 3
2 2
= 23=2▯

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