જો int_{0}^{100 pi} frac{sin ^{2} x}{e^{left( | vedclass

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(B) ધારો કે $I = \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x$. વિધેય $f(x) = \frac{\sin^2 x}{

Vedclass · Accepted Answer

$200(1-e^{-1})$

Vedclass · Answer

(B) ધારો કે $I = \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x$. વિધેય $f(x) = \frac{\sin^2 x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}$ એ $\pi$ આવર્તકાળ ધરાવે છે,તેથી $I = 100 \int_{0}^{\pi} \frac{\sin^2 x}{e^{x/\pi}} dx$. $I = 100 \int_{0}^{\pi} e^{-x/\pi} \left(\frac{1-\cos 2x}{2}\right) dx = 50 \left[ \int_{0}^{\pi} e^{-x/\pi} dx - \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx \right]$. ધારો કે $I_1 = \int_{0}^{\pi} e^{-x/\pi} dx = [-\pi e^{-x/\pi}]_0^{\pi} = \pi(1-e^{-1})$. ધારો કે $I_2 = \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$I_2 = [-\pi e^{-x/\pi} \cos 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \sin 2x) dx = \pi(1-e^{-1}) - 2\pi \int_0^{\pi} e^{-x/\pi} \sin 2x dx$. બીજા ભાગનું ફરીથી સંકલન કરતા,$\int_0^{\pi} e^{-x/\pi} \sin 2x dx = [-\pi e^{-x/\pi} \sin 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \cos 2x) dx = 2\pi I_2$. આમ,$I_2 = \pi(1-e^{-1}) - 2\pi(2\pi I_2) = \pi(1-e^{-1}) - 4\pi^2 I_2$. $I_2(1+4\pi^2) = \pi(1-e^{-1}) \implies I_2 = \frac{\pi(1-e^{-1})}{1+4\pi^2}$. કિંમત મૂકતા,$I = 50 [\pi(1-e^{-1}) - \frac{\pi(1-e^{-1})}{1+4\pi^2}] = 50 \pi(1-e^{-1}) [\frac{4\pi^2}{1+4\pi^2}] = \frac{200(1-e^{-1})\pi^3}{1+4\pi^2}$. $\frac{\alpha \pi^3}{1+4\pi^2}$ સાથે સરખાવતા,$\alpha = 200(1-e^{-1})$ મળે છે.

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