Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.4 Exercises: 90

Answer

$F^{\prime}(x)=2x^{-5}$

Work Step by Step

$F(x)=\displaystyle \int_{2}^{x^{2}}\frac{1}{t^{3}}dt=\int_{2}^{x^{2}}t^{-3}dt$ $=[\displaystyle \frac{1}{-2}\cdot t^{-2}]_{2}^{x^{2}}$ $=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{2\cdot 2^{4}}$ $=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{8}$ $F(x)=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{8}$ $F^{\prime}(x)=-\displaystyle \frac{1}{2}(-4x^{-5})=2x^{-5}$
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