Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 337: 28

Answer

$2\frac{7}{24} $

Work Step by Step

First, we expand the terms being squared and get the polynomial $4 - \frac{4}{p^2} + \frac{1}{p^4} $ as our integrand. We can rewrite the terms to get rid of the fractions and get $4 - 4p^{-2} + {p^{-4}} $ Evaluating this integral gives us $4p - (-)4p^{-1} +\frac{p^{-3}}{-3} = 4p + \frac{4}{p} -\frac{1}{3p^3}$ We now substitute our bounds of integration $2$ and $1$ and get $(4(2) + \frac{4}{2} - \frac{1}{3(2)^3}) - (4 + 4 - \frac{1}{3}) = (8 + 2 - \frac{1}{24}) - (8 - \frac{1}{3}) = 2 - \frac{1}{24} + \frac{8}{24} = 2\frac{7}{24} $
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