int_1^2 frac{1}{x^2} e^{-frac{1}{x}} , dx = | vedclass

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(D) Let $I = \int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx$. Substitute $t = -\frac{1}{x}$. Then,$dt = \frac{1}{x^2} \, dx$. When $x = 1$,$t = -1$. Wh

Vedclass · Accepted Answer

$\frac{e^{1/2} - 1}{e}$

Vedclass · Answer

(D) Let $I = \int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx$. Substitute $t = -\frac{1}{x}$. Then,$dt = \frac{1}{x^2} \, dx$. When $x = 1$,$t = -1$. When $x = 2$,$t = -\frac{1}{2}$. Substituting these into the integral: $I = \int_{-1}^{-1/2} e^t \, dt = [e^t]_{-1}^{-1/2}$. $I = e^{-1/2} - e^{-1} = \frac{1}{\sqrt{e}} - \frac{1}{e}$. $I = \frac{\sqrt{e} - 1}{e}$.

$\int_1^2 \frac{1}{x^2} e^{-\frac{1}{x}} \, dx = $

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