Calculus: Early Transcendentals 8th Edition

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: 1

Answer

$f_{avg}=7$

Work Step by Step

Given: Average value of the function $f(x)=3x^{2}+8x$ over the interval $[-1,2]$ is given by Average value of the function $f(x)$ over the interval $[a.b]$ is given by $f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$ $f_{avg}=\frac{1}{2-(-1)}\int_{-1}^{2}3x^{2}+8xdx$ $=\frac{1}{3}[x^{3}+4x^{2}]_{-1}^{2}$ $=\frac{1}{3}[8+16]-\frac{1}{3}[-1+4]$ $=7$ Hence, $f_{avg}=7$
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