Answer
$c$=$\frac{3}{125}$
Work Step by Step
$\int^5_0cx\sqrt {25-x^2}dx$
=$-\frac{1}{3}c(25-x^2)^\frac{3}{2}]^5_0$
=$\frac{125}{3}c$
so,
$c$=$\frac{3}{125}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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: 20
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