-12 e x 1- x -1- x 2 dx -A - e 2-2- ee 2B- e-e2-2C- e 2-2- eD- N one of these

The correct option is C e^2/2 e∫_1^2 e^x ( 1/x 1/x^2 )dx= [ 1/xe^x ]_1^2=e^2/2 e ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫12 e x 1/ x ...

Question

$\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x =$

\frac{e^{2}}{2} + e

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e - \frac{e^{2}}{2}

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\frac{e^{2}}{2} - e

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Solution

The correct option is C $\frac{e^{2}}{2} - e$
$\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x = {[\frac{1}{x} e^{x}]}_{1}^{x} = \frac{e^{2}}{2} - e$

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Similar questions

\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x =

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\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x =

\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x

equals to

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∫21ex(1x−1x2)dx=

The correct option is C e22−e∫21ex(1x−1x2)dx=[1xex]21=e22−e

$\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x =$

The correct option is C $\frac{e^{2}}{2} - e$
$\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x = {[\frac{1}{x} e^{x}]}_{1}^{x} = \frac{e^{2}}{2} - e$