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$\int_{0}^{1} x(1-x)^{3 / 2} d x$ is
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The correct answer is:
$\frac{4}{35}$
$\int_{0}^{1} x(1-x)^{3 / 2} d x$
$=\int_{0}^{1} \mathrm{x}^{(2-1)}(1-\mathrm{x})^{5 / 2-1} \mathrm{dx}$
$=\mathrm{B}(2,5 / 2)\left\{\begin{array}{l}\because \mathrm{B}(\mathrm{m}, \mathrm{n})=\int_{0}^{1} \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ \because \mathrm{B}(\mathrm{m}, \mathrm{n})=\frac{\Gamma \mathrm{m} \Gamma \mathrm{n}}{\Gamma(\mathrm{m}+\mathrm{n})}\end{array}\right.$
$=\frac{\Gamma 2 \Gamma 5 / 2}{\Gamma(2+5 / 2)}$
$=\frac{1 \cdot 3 / 2 \cdot 1 / 2 \Gamma 1 / 2}{\Gamma 9 / 2} \quad(\because \Gamma 1 / 2=\sqrt{\pi})$
$=\frac{3 / 4 \cdot \sqrt{\pi}}{7 / 2 \cdot 5 / 2 \cdot 3 / 2 \cdot 1 / 2 \sqrt{\pi}}=\frac{4}{35}$
$=\int_{0}^{1} \mathrm{x}^{(2-1)}(1-\mathrm{x})^{5 / 2-1} \mathrm{dx}$
$=\mathrm{B}(2,5 / 2)\left\{\begin{array}{l}\because \mathrm{B}(\mathrm{m}, \mathrm{n})=\int_{0}^{1} \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx} \\ \because \mathrm{B}(\mathrm{m}, \mathrm{n})=\frac{\Gamma \mathrm{m} \Gamma \mathrm{n}}{\Gamma(\mathrm{m}+\mathrm{n})}\end{array}\right.$
$=\frac{\Gamma 2 \Gamma 5 / 2}{\Gamma(2+5 / 2)}$
$=\frac{1 \cdot 3 / 2 \cdot 1 / 2 \Gamma 1 / 2}{\Gamma 9 / 2} \quad(\because \Gamma 1 / 2=\sqrt{\pi})$
$=\frac{3 / 4 \cdot \sqrt{\pi}}{7 / 2 \cdot 5 / 2 \cdot 3 / 2 \cdot 1 / 2 \sqrt{\pi}}=\frac{4}{35}$
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