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$\int_{0}^{1} x(1-x)^{5} d x=$
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$\frac{1}{42}$
(C)
$\begin{aligned} \text { Let } I &=\int_{0}^{1} x(1-x)^{5} d x \\ \therefore I &=\int_{0}^{1}(1-x)(1-(1-x))^{5} d x=\int_{0}^{1}(1-x) x^{5} d x=\int\left(x^{5}-x^{6}\right) d x \\ &=\left[\frac{x^{6}}{6}-\frac{x^{7}}{7}\right]_{0}^{1}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42} \end{aligned}$
$\begin{aligned} \text { Let } I &=\int_{0}^{1} x(1-x)^{5} d x \\ \therefore I &=\int_{0}^{1}(1-x)(1-(1-x))^{5} d x=\int_{0}^{1}(1-x) x^{5} d x=\int\left(x^{5}-x^{6}\right) d x \\ &=\left[\frac{x^{6}}{6}-\frac{x^{7}}{7}\right]_{0}^{1}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42} \end{aligned}$
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