int_{frac{1}{25}}^3 frac{e^{frac{3}{x}}}{x^2} | vedclass

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(D) Let $I = \int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x$. Substitute $t = \frac{3}{x}$. Then $dt = -\frac{3}{x^2} dx$,which implies $\frac{

Vedclass · Accepted Answer

$\frac{1}{3}(e^{75}-e)$

Vedclass · Answer

(D) Let $I = \int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x$. Substitute $t = \frac{3}{x}$. Then $dt = -\frac{3}{x^2} dx$,which implies $\frac{1}{x^2} dx = -\frac{1}{3} dt$. When $x = 3$,$t = \frac{3}{3} = 1$. When $x = \frac{1}{25}$,$t = \frac{3}{1/25} = 75$. Substituting these into the integral: $I = \int_{75}^1 e^t \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int_{75}^1 e^t dt$. $I = -\frac{1}{3} [e^t]_{75}^1 = -\frac{1}{3} (e^1 - e^{75})$. $I = \frac{1}{3} (e^{75} - e)$.

$\int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x=$

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