Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 276: 60

Answer

$\frac{3}{2}$

Work Step by Step

av(f)=$(\frac{1}{1-(-2)})\int^{1}_{-2}(t^2-t)dt$ =$\frac{1}{3}\int^{1}_{-2}t^2dt-\frac{1}{3}\int^{1}_{-2}tdt$ =$\frac{1}{3}\int^{1}_{0}t^2dt-\frac{1}{3}\int^{-2}_{0}t^2dt$ =$\frac{1}{3}(\frac{1}{2}-\frac{4}{2})$ =$\frac{1}{3}$($\frac{1^3}{3}$)-$\frac{1}{3}$($\frac{-2^3}{3}$)+$\frac{1}{3}=\frac{3}{2}$
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