Chapter 5 - Applications of Integration - 5.5 Average Value of a Function - 5.5 Exercises - Page 391: 6

${1\over 24}$

$f_{ave} = {1\over(b-a)} \int^b_a f(x) dx = {1\over(1-(-1))} \int ^1_{-1} {(x^2)\over(x^3 + 3)^2} dx = {1\over2} \int^4_2 {1\over {u^2}}({1\over 3}du) [u = x^3 +3]$ ${1\over 6} [-{1 \over u}]^4_2 = {1 \over 6} (-{1 \over 4} + {1\over 2}) = {1\over 24}$