where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is :
$100 \int_{0}^{\pi} e^{-x / \pi} \frac{(1-\cos 2 x)}{2} d x$
$=50\left\{\int_{0}^{\pi} e^{-x / \pi} d x-\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x\right\}$
$I_{1}=\int_{0}^{\pi} e^{-x / \pi} d x=\left[-\pi e^{-x / \pi}\right]_{0}^{\pi}=\pi\left(1-e^{-1}\right)$
$I_{2}=\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x$
$\left.=-\pi \mathrm{e}^{-x / \pi} \cos 2 x\right]_{0}^{\pi}-\int-\pi e^{-x / \pi}(-2 \sin 2 x) d x$
$=\pi\left(1-e^{-1}\right)-2 \pi \int_{0}^{\pi} e^{-x / \pi} \sin 2 x d x$
$\left.=\pi\left(1-e^{-1}\right)-2 \pi\left\{-\pi e^{-x / \pi} \sin 2 x\right]_{0}^{\pi}-\int_{0}^{\pi}-\pi e^{-x / \pi} 2 \cos 2 x d x\right\}$
$=\pi\left(1-\mathrm{e}^{-1}\right)-4 \pi^{2} I_{2}$
$\Rightarrow I_{2}=\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}$
$\therefore I=50\left\{\pi\left(1-e^{-1}\right)-\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}\right\}$
$=\frac{200\left(1-e^{-1}\right) \pi^{3}}{1+4 \pi^{2}}$
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$\frac{2}{\sqrt{26}}$
$f(x)= \begin{cases}n(1-2 n x) \text { if } 0 \leq x \leq \frac{1}{2 n}2 n(2 n x-1) \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n}4 n(1-n x) \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(n x-1) \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$ , then the maximum value of the function $f$ is