Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.9 - Probability - Exercises 8.9 - Page 514: 17


$c$=$\sqrt {21}$

Work Step by Step

$\int^c_3\frac{1}{6}xdx$ =$\frac{1}{12}c^2-\frac{3}{4}$ solving $\frac{1}{12}c^2-\frac{3}{4}$ =$1$ we find $c$=$\sqrt {21}$
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