Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 320: 14

Answer

$$\frac{{31}}{{160}}$$

Work Step by Step

$$\eqalign{ & \int_1^2 {\frac{1}{{{x^6}}}} dx \cr & {\text{negative exponent}} \cr & = \int_1^2 {{x^{ - 6}}} dx \cr & {\text{find the antiderivative by the power rule}} \cr & = \left( {\frac{{{x^{ - 5}}}}{{ - 5}}} \right)_1^2 \cr & = - \frac{1}{5}\left( {\frac{1}{{{x^5}}}} \right)_1^2 \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = - \frac{1}{5}\left( {\frac{1}{{{2^5}}} - \frac{1}{{{1^5}}}} \right) \cr & {\text{simplify}} \cr & = - \frac{1}{5}\left( { - \frac{{31}}{{32}}} \right) \cr & = \frac{{31}}{{160}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.