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Section 8.4 Improper Integrals 461<br />
Thus, A B 0 and 3A B 2. Solving, we find A 1 and B 1. Therefore,<br />
2 1 1<br />
x 2 <br />
4x 3 x 1 x 3<br />
and<br />
b<br />
b<br />
b<br />
2 dx<br />
dx<br />
dx<br />
0 x 2 4x<br />
3 0 x<br />
<br />
1 0 x 3<br />
b<br />
b<br />
ln (x 1) ln (x 3) <br />
So,<br />
Thus,<br />
<br />
lim<br />
b→ ln b 1<br />
b 3<br />
<br />
ln 3<br />
lim<br />
0<br />
ln (b 1) ln (b 3) ln 3<br />
ln b 1<br />
ln 3<br />
b 3<br />
b→ ln 1 1<br />
b<br />
1 3b<br />
<br />
ln 3<br />
<br />
0<br />
ln 3.<br />
2 dx<br />
<br />
0 x 2 ln 3. Now try Exercise 13.<br />
4x<br />
3<br />
In Example 4 we use l’Hôpital’s Rule to help evaluate the improper integral.<br />
EXAMPLE 4 Using L’Hôpital’s Rule with Improper Integrals<br />
Evaluate <br />
1 xex dx or state that it diverges.<br />
SOLUTION<br />
By definition <br />
1 xex dx lim<br />
b→<br />
b 1 xex dx. We use integration by parts to evaluate the<br />
definite integral. Let<br />
u x dv e x dx<br />
du dx v e x .<br />
Then<br />
So,<br />
b<br />
1<br />
b b<br />
e x dx<br />
1 1<br />
<br />
[ x]<br />
b<br />
xex e<br />
1<br />
b<br />
<br />
[<br />
(x 1)ex]<br />
1<br />
xe x dx <br />
[ xex]<br />
(b 1)e b 2e 1 .<br />
lim [(b <br />
b→ 1)eb 2e 1 ] lim (b 1)<br />
<br />
b→ eb<br />
2 e <br />
lim 1 2 b→ e b e <br />
2 e .<br />
Thus <br />
1 xex dx 2/e. Now try Exercise 17.
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