Chapter 7 - Techniques of Integration - 7.8 Improper Integrals - 7.8 Exercises - Page 574: 3

T = $\infty$........... A =0.5

$\int1/x^{3}$ = $-x^{-2}/2$ = $-1/(2x^{2})$ $-1/(2*10^{2})$ - $-1/(2*1^{2})$ = 0.495 $-1/(2*100^{2})$ - $-1/(2*1^{2})$ = 0.49995 $-1/(2*1000^{2})$ - $-1/(2*1^{2})$ = 0.4999995 as x approaches infinity the number gets closer and closer to 0.5 meaning $\lim\limits_{x \to \infty}$ $-1/(2*x^{2})$ - $-1/(2*1^{2})$ = 0.5