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  • If <img alt="\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in \mathbf{R}" src="/latex-image/?%5Cint_%7B0%7D%5E%7B100%20%5Cpi%7D%20%5Cfrac%7B%5Csin%20%5E%7B2

If \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in \mathbf{R}where [x] is the greatest integer less than or equal to x, then the value of \alpha is :
 
Option: 1 200\left(1-e^{-1}\right)
Option: 2 100(1-\mathrm{e})
Option: 3 50(e-1)
Option: 4 150\left(\mathrm{e}^{-1}-1\right)

Answers (1)

best_answer

I=\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left\{\frac{x}{\pi}\right\}}}
Now e^{\{x\}}$ is periodic with period $=1

\Rightarrow e^{\left\{\frac{x}{\pi}\right\}} is periodic with period =\frac{1}{1 / \pi}=\pi.

\begin{aligned} \therefore I &=100 \int_{0}^{\pi} \frac{\sin ^{2} x}{e^{x / \pi}} d x \\ &=100\left[\int_{0}^{\pi} e^{-x / \pi}\left(\frac{1-\cos 2 x}{2}\right) d x\right] \\ &=50\left[\int_{0}^{\pi} e^{-x / \pi}-\int_{0}^{\pi} e^{-x / \pi} \cos 2 x d x\right] \end{aligned}

         =50\left [ I_{1}-I_{2} \right ]--------(i)

\begin{aligned} &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 \text {. }\\ &\text { Using integration by parts' }\\ &I_{2}=\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}\\ &\therefore(i) \Rightarrow I=50\left(\pi\left(1-e^{-1}\right)-\frac{\pi\left(1-e^{-1}\right)}{1+4 \pi^{2}}\right)\\ \end{aligned}

                     \begin{aligned} &=\frac{200\left(1-e^{-1}\right) \pi^{3}}{1+4 \pi^{2}}\\ \end{aligned}

\begin{aligned} &\text { comparing } \alpha=200\left(1-e^{-1}\right)\end{aligned}

Hence option (1) is correct.

Posted by

Kuldeep Maurya

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