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Section 8.4 Improper Integrals 463<br />
DEFINITION Improper Integrals with Infinite Discontinuities<br />
Integrals of functions that become infinite at a point within the interval of integration<br />
are improper integrals.<br />
1. If f x is continuous on a, b, then<br />
b<br />
f x dx lim<br />
c→a b<br />
f x dx.<br />
a<br />
c<br />
2. If f x is continuous on a, b, then<br />
b<br />
f x dx lim<br />
c→b c<br />
f x dx.<br />
a<br />
a<br />
3. If f x is continuous on a, c c, b, then<br />
f x dx c<br />
f x dx b<br />
f x dx.<br />
b<br />
a<br />
a<br />
c<br />
In parts 1 and 2, if the limit is finite the improper integral converges and the limit is the<br />
value of the improper integral. If the limit fails to exist the improper integral diverges. In<br />
part 3, the integral on the left-hand side of the equation converges if both integrals on the<br />
right-hand side have values, otherwise it diverges.<br />
EXPLORATION 1<br />
1<br />
Investigation 0<br />
d x<br />
xp<br />
1. Explain why these integrals are improper if p 0.<br />
2. Show that the integral diverges if p 1.<br />
3. Show that the integral diverges if p 1.<br />
4. Show that the integral converges if 0 p 1.<br />
EXAMPLE 6<br />
Evaluate 3<br />
SOLUTION<br />
0<br />
Infinite Discontinuity at an Interior Point<br />
dx<br />
.<br />
x 1 23<br />
The integrand has a vertical asymptote at x 1 and is continuous on 0, 1 and 1, 3.<br />
Thus, by part 3 of the definition above<br />
3<br />
0<br />
x <br />
dx<br />
1 23<br />
1<br />
0<br />
x <br />
dx<br />
1 23<br />
3<br />
1<br />
dx<br />
.<br />
x 1 23<br />
Next, we evaluate each improper integral on the right-hand side of this equation.<br />
1<br />
0<br />
dx<br />
x 1 23 lim<br />
c→1 c<br />
lim<br />
c→1 <br />
dx<br />
<br />
x 1 23<br />
0<br />
c<br />
3x 113]<br />
0<br />
lim<br />
c→1 3c 113 3 3<br />
continued
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