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,
$c$=$\frac{3}{125}$
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