University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 342: 15



Work Step by Step

We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ where $C$ is a constant of proportionality. Now, $\int_1^2(x-\dfrac{1}{x^2}) dx=[\dfrac{x^{1+1}}{1+1}-\dfrac{x^{-2+1}}{-2+1}]_1^2$ or, $[\dfrac{x^2}{2}+\dfrac{1}{x}]_1^2=2+1/2-1/2-1=1$
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