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: 35

Answer

$\displaystyle\int\limits_1^2\dfrac{v^{3}+3v^{6}}{v^{4}}dv=\ln2+ 7$

Work Step by Step

$\displaystyle\int\limits_1^2\dfrac{v^{3}+3v^{6}}{v^{4}}dv$ Take the denominator up to multiply the numerator by changing the sign of its exponent: $\displaystyle\int\limits_1^2(v^{3}+3v^{6})v^{-4}dv=\int\limits_1^2(v^{-1}+3v^{2})dv=\int\limits_1^2\Big(\dfrac{1}{v}+3v^{2}\Big)dv=...$ Integrate each term separately and apply the second part of the fundamental theorem of calculus: $\displaystyle\int\limits_1^2\dfrac{1}{v}dv+3\int\limits_1^2v^{2}dv=\ln v+3\Big(\dfrac{1}{3}\Big)v^{3}\Big|_1^2=\ln v+v^{3}\Big|_1^2=...$ $...=[\ln2+(2)^{3}]-[\ln1+(1)^{3}]=\ln2+8-0-1=\ln2+ 7$
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