Evaluate the following integrals : $\int_0^a x^2(a-x)^{3 / 2} d x$
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Answer
We use the property $ \begin{aligned} & \int_0^a f(x) d x=\int_0^a f(a-x) d x \\ & \therefore \int_0^a x^2(a-x)^{\frac{3}{2}} d x=\int_0^a(a-x)^2(a-a+x)^{\frac{3}{2}} d x \\ & =\int_0^a\left(a^2-2 a x+x^2\right) x^{\frac{3}{2}} d x \\ & =\int_0^a\left(a^2 x^{\frac{3}{2}}-2 a x^{\frac{5}{2}}+x^{\frac{7}{2}}\right) d x \\ & =a^2 \int_0^a x^{\frac{3}{2}} d x-2 a \int_0^a x^{\frac{5}{2}} d x+\int_0^a x^{\frac{7}{2}} d x \\ & =a^2\left[\frac{x^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}\right]_0^a-2 a\left[\frac{x^{\frac{7}{2}}}{\left(\frac{7}{2}\right)}\right]_0^a+\left[\frac{x^{\frac{9}{2}}}{\left(\frac{9}{2}\right)}\right]_0^a \\ & =\frac{2 a^2}{5}\left[a^{\frac{5}{2}}-0\right]-\frac{4 a}{7}\left[a^{\frac{7}{2}}-0\right]+\frac{2}{9}\left[a^{\frac{9}{2}}-0\right] \\ & =\frac{2}{5} a^{\frac{9}{2}}-\frac{4}{7} a^{\frac{9}{2}}+\frac{2}{9} a^{\frac{9}{2}}=\left(\frac{2}{5}-\frac{4}{7}+\frac{2}{9}\right) a^{\frac{9}{2}} \\ & =\left(\frac{126-180+70}{315}\right) a^{\frac{9}{2}}=\frac{16}{315} a^{\frac{9}{2}} . \end{aligned} $
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