Answer
$-2$
Work Step by Step
We know that $av(f)=\dfrac{1}{b-a}\int_{a}^{b} f(x) dx$
Therefore,
$av(f)=\dfrac{1}{1-0} \int_{0}^{1} (-3x^2-1) \ dx \\=(-3) \int_0^1 x^2 \ dx-\int_0^1 (1) \ dx\\=(-3) [\dfrac{1}{3}-(1-0)]\\=-2$
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: 57
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