Answer
=$0.688$
Work Step by Step
$\int^1_0\frac{3}{2}x(2-x)dx$
=$-\frac{1}{2}x^2(x-3)]^1_\frac{1}{2}$
=$\frac{11}{16}$=$0.688$
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: 13
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