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: 27

Answer

$\displaystyle\int\limits_0^1(u+2)(u-3)du=-\dfrac{37}{6}$

Work Step by Step

$\displaystyle\int\limits_0^1(u+2)(u-3)du$ Evaluate the product present in the integral: $\displaystyle\int\limits_0^1(u^{2}-3u+2u-6)du=\displaystyle\int\limits_0^1(u^{2}-u-6)du=...$ Integrate each term separately and apply the second part of the fundamental theorem of calculus: $\displaystyle\int\limits_0^1u^{2}du-\int\limits_0^1udu-6\int\limits_0^1du=\dfrac{1}{3}u^{3}-\dfrac{1}{2}u^{2}-6u\Big|_0^1=...$ $...=\Big[\dfrac{1}{3}(1)^{3}-\dfrac{1}{2}(1)^{2}-6(1)\Big]-\Big[\dfrac{1}{3}(0)^{3}-\dfrac{1}{2}(0)^{2}-6(0)\Big]=...$ $...=\dfrac{1}{3}-\dfrac{1}{2}-6=-\dfrac{37}{6}$
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