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 - Page 463: 4

Answer

$\sqrt{3}-1$

Work Step by Step

$$ g(t)=\frac{t}{\sqrt{3+t^{2}}},[1,3] $$ The average value of the given function on the given interval is $$ \begin{split} g_{\text {ave }}&=\frac{1}{b-a} \int_{a}^{b} g(t) d t \\ &=\frac{1}{3-1} \int_{1}^{3} \frac{t}{\sqrt{3+t^{2}}} d t \\ &=\frac{1}{2}\left[\left(3+t^{2}\right)^{1 / 2}\right]_{1}^{3} \\ &=\frac{1}{2}(2 \sqrt{3}-2) \\ &=\sqrt{3}-1 \end{split} $$
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