Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Practice Exercises - Page 308: 45



Work Step by Step

Integrate with respect to x by taking the antiderivative of each term. Then plug in limits to evaluate: $=\frac{3x^3}{3}-\frac{4x^2}{2}+7x=x^3-2x^2+7x$ with limits from -1 to 1 $=[(1)^3-2(1)^2+7(1)]-[(-1)^3-2(-1)^2+7(-1)]$ $=(1-2+7)-(-1-2-7)=6-(-10)=16$
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