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: 22

Answer

$\displaystyle\int\limits_0^1(1-8v^{3}+16v^{7})dv=1$

Work Step by Step

$\displaystyle\int\limits_0^1(1-8v^{3}+16v^{7})dv$ Integrate each term separately: $\displaystyle\int\limits_0^1dv-\int\limits_0^18v^{3}dv+\int\limits_0^116v^{7}dv=...$ Take the constants out of the integrals and continue with the process: $...=\displaystyle\int\limits_0^1dv-8\int\limits_0^1v^{3}dv+16\int\limits_0^1v^{7}dv=...$ $...=v-8(\dfrac{1}{4})v^{4}+16(\dfrac{1}{8})v^{8}\Big|_0^1=v-2v^{4}+2v^{8}\Big|_0^1=...$ Apply the fundamental theorem of calculus to get the answer: $...=\Big[1-2(1)^{4}+2(1)^{8}\Big]-\Big[0-2(0)^{4}+2(0)^{8}\Big]=...$ $...=1-2+2=1$
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