Improper Integrals Involving Infinity:
We define
Example:
int from 1 to infinity of 1/x2 dx = lim as m
-> infinity of int from a to m of 1/x2 dx
= lim as m -> infinity of -1/x from 1 to infinity = lim
as m -> infinity of -1/m - (-1/1) = 0 + 1 = 1
Exercises:
A) int from 0 to infinity of xe-x dx
B) int from -infinity to infinity of 1/(1 +
x2) dx
C) int from 0 to infinity of sinx dx
D) Determine for which values of p the integral
int from 1 to infinity of 1/xp dx converges.
E) Use the formula for arclength to show that the circumference of
the semi-circle y = sqrt(r2 - x2) is (pi)(r).