## Thomas' Calculus 13th Edition

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