Answer
$0$
Work Step by Step
Since $av(f)=\dfrac{1}{B-A}\int_{A}^{B} f(x) dx$
Here, we have
$av(f)=\dfrac{1}{\sqrt 3-0}\int_{0}^{\sqrt 3} (x^2-1) dx$
This implies that
$\dfrac{1}{\sqrt 3-0}[\dfrac{(\sqrt 3)^3-(0)^3}{3}-(\sqrt 3-0)]=0$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 55
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