Question
Evaluate: $\int_0^3 x^2(3-x)^{\frac{5}{2}} d x$

Answer

$\text { Let } I =\int_0^3 x^2(3-x)^{\frac{5}{2}} d x$
$=\int_0^3(3-x)^2[3-(3-x)]^{\frac{5}{2}} d x \quad \ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^3\left(9-6 x+x^2\right) x^{\frac{5}{2}} d x$
$=\int_0^3\left(9 x^{\frac{5}{2}}-6 x^{\frac{7}{2}}+x^{\frac{9}{2}}\right) d x$
$=9 \int_0^2 x^{\frac{5}{2}} d x-6 \int_0^3 x^{\frac{7}{2}} d x+\int_0^3 x^{\frac{9}{2}} d x$
$=9\left[\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right]_0^3-6\left[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}\right]_0^3+\left[\frac{x^{\frac{11}{2}}}{\frac{11}{2}}\right]_0^3$
$=\frac{18}{7}\left[(3)^{\frac{7}{2}}-0\right]-\frac{12}{9}\left[(3)^{\frac{9}{2}}-0\right]+\frac{2}{1}\left[(3)^{\frac{11}{2}}-0\right]$
$=\left[\frac{18}{7}-\left(\frac{12}{9} \times 3\right)+\left(\frac{2}{11} \times 9\right)\right](3)^{\frac{7}{2}}$
$=\left(\frac{198-308+126}{77}\right)(3)^{\frac{7}{2}}$
$\therefore I =\frac{16}{77}(3)^{\frac{7}{2}}$

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