Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 30

Answer

$\displaystyle\int\limits_{-1}^2(3u-2)(u+1)du=\dfrac{9}{2}$

Work Step by Step

$\displaystyle\int\limits_{-1}^2(3u-2)(u+1)du$ Evaluate the product inside the integral: $\displaystyle\int\limits_{-1}^2(3u^{2}+3u-2u-2)du=\int\limits_{-1}^2(3u^{2}+u-2)du=...$ Integrate each term separately and apply the second part of the fundamental theorem of calculus: $...=\displaystyle3\int\limits_{-1}^2u^{2}du+\int\limits_{-1}^2udu-2\int\limits_{-1}^2du=3\Big(\dfrac{1}{3}\Big)u^{3}+\dfrac{1}{2}u^{2}-2u\Big|_{-1}^2=...$ $...=u^{3}+\dfrac{1}{2}u^{2}-2u\Big|_{-1}^2=...$ $...=\Big[(2)^{3}+\dfrac{1}{2}(2)^{2}-2(2)\Big]-\Big[(-1)^{3}+\dfrac{1}{2}(-1)^{2}-2(-1)\Big]=...$ $...=8+\dfrac{1}{2}(4)-4-\Big[-1+\dfrac{1}{2}+2\Big]=8+2-4+1-\dfrac{1}{2}-2=$ $...=\dfrac{9}{2}$
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