Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 19



Work Step by Step

We have $$\int_{1/2}^{1} 8x^{-3}dx=\frac{1}{-2}8x^{-2}|_{1/2}^1=-4x^{-2}|_{1/2}^1=-4+16=12.$$
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