Answer
$\frac{13}{6}$
Work Step by Step
the average value is
= $\frac{1}{3-0}\int_0^{3}x(x)dx$
= $\frac{1}{3}(\int_0^{1}x·0dx+\int_1^{2}x·1dx+\int_2^{3}x·2dx)$
= $\frac{1}{3}(\frac{1}{2}x^{2}|_1^2+x^{2}|_2^3)$
= $\frac{13}{6}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 24
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