Answer
$\frac{1}{24}$
Work Step by Step
$f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$
In our case, $a=1$ and $b=-1$
So $f_{avg}=\frac{1}{2}\int_{-1}^{1}\frac{x^2}{(x^3+3)^2}dx$
$=-\frac{1}{6}\left[\frac{1}{x^3+3}\right]_{-1}^{1}$
$=\frac{1}{24}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 6 - Section 6.5 - Average Value of a Function - 6.5 Exercises - Page 463: 6
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