The value of the integral int_{1/pi }^{2/pi } | vedclass

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(D) Let $t = \frac{1}{x}$.Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.When $x = \frac{1}{\pi}$,$t = \pi$.When $x = \fra

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$1$

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(D) Let $t = \frac{1}{x}$.Then,$dt = -\frac{1}{x^2} \,dx$,which implies $-\,dt = \frac{1}{x^2} \,dx$.When $x = \frac{1}{\pi}$,$t = \pi$.When $x = \frac{2}{\pi}$,$t = \frac{\pi}{2}$.Substituting these into the integral:$\int_{1/\pi}^{2/\pi} \frac{\sin(1/x)}{x^2} \,dx = \int_{\pi}^{\pi/2} \sin(t) \cdot (-dt)$$= \int_{\pi/2}^{\pi} \sin(t) \,dt$$= [-\cos(t)]_{\pi/2}^{\pi}$$= -(\cos(\pi) - \cos(\pi/2))$$= -(-1 - 0) = 1$.

The value of the integral $\int_{1/\pi }^{2/\pi } \frac{\sin(1/x)}{x^2} \,dx$ is:

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