Answer
$$ 0.4\rm L$$
Work Step by Step
$$V_{ave} = \frac{1}{5} \int^5_0 V(t) dt = \frac{1}{4\pi} \int^5_{0} [1-cos(\frac{2}{5} \pi t)] dt $$
$$=\frac{1}{4\pi} [t - \frac{5}{2\pi} sin (\frac{2}{5} \pi t)]^5_0 = \frac{5}{4\pi} \approx 0.4\rm L$$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 5 - Applications of Integration - 5.5 Average Value of a Function - 5.5 Exercises - Page 391: 21
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