Let f(alpha) = int_{0}^{alpha} x^{2} left(1 - | vedclass

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(A) Given $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$. Let $t = 1 - \frac{x}{\alpha}$,then $x = \alpha(1-t)$ a

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$\frac{25}{168}$

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(A) Given $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$. Let $t = 1 - \frac{x}{\alpha}$,then $x = \alpha(1-t)$ and $dx = -\alpha dt$. When $x=0, t=1$ and when $x=\alpha, t=0$. $f(\alpha) = \int_{1}^{0} \alpha^{2}(1-t)^{2} t^{\alpha} (-\alpha dt) = \alpha^{3} \int_{0}^{1} (1-2t+t^{2}) t^{\alpha} dt$. $f(\alpha) = \alpha^{3} \int_{0}^{1} (t^{\alpha} - 2t^{\alpha+1} + t^{\alpha+2}) dt$. $f(\alpha) = \alpha^{3} \left[ \frac{t^{\alpha+1}}{\alpha+1} - \frac{2t^{\alpha+2}}{\alpha+2} + \frac{t^{\alpha+3}}{\alpha+3} \right]_{0}^{1} = \alpha^{3} \left( \frac{1}{\alpha+1} - \frac{2}{\alpha+2} + \frac{1}{\alpha+3} \right)$. Thus,$\frac{f(\alpha)}{\alpha^{3}} = \left( \frac{1}{\alpha+1} - \frac{1}{\alpha+2} \right) - \left( \frac{1}{\alpha+2} - \frac{1}{\alpha+3} \right)$. Summing from $\alpha=1$ to $5$: $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}} = \sum_{\alpha=1}^{5} \left[ \left( \frac{1}{\alpha+1} - \frac{1}{\alpha+2} \right) - \left( \frac{1}{\alpha+2} - \frac{1}{\alpha+3} \right) \right]$. This is a telescoping sum: $\left( \frac{1}{2} - \frac{1}{3} - \frac{1}{3} + \frac{1}{4} \right) + \dots + \left( \frac{1}{6} - \frac{1}{7} - \frac{1}{7} + \frac{1}{8} \right)$. $= \left( \frac{1}{2} - \frac{1}{3} \right) - \left( \frac{1}{7} - \frac{1}{8} \right) = \frac{1}{6} - \frac{1}{56} = \frac{28-3}{168} = \frac{25}{168}$.

Let $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$ (where $\alpha > 0$),then $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}}$ is equal to-

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