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: 22


=$ 0.353$

Work Step by Step

$\int^1_0c\sqrt x(1-x)dx$ =$c(\frac{2}{3}x^\frac{3}{2}-\frac{2}{5}x^\frac{5}{2})]^1_0$ =$\frac{4}{15}c$, So, $c$=$\frac{15}{4}$. Then $\int^\frac{1}{2}_\frac{1}{4}\frac{15}{4}\sqrt x(1-x)dx$ =$\frac{7}{16}\sqrt 2-\frac{17}{64}$ =$0.353$
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