A) \[e\left( \frac{e}{2}-1 \right)\]
B) \[e(e-1)\]
C) \[e-\frac{1}{e}\]
D) 0
Correct Answer: A
Solution :
[a] Let \[I=\int_{1}^{2}{{{e}^{x}}\left( \frac{1}{x}-\frac{1}{{{x}^{2}}} \right)dx}\] \[=\int\limits_{1}^{2}{{{e}^{x}}(f(x)+f'(x))dx}\] where \[f(x)=\frac{1}{x}\] \[={{e}^{x}}f\left. (x) \right|_{1}^{2}\] \[\therefore I=\left. \frac{{{e}^{x}}}{x} \right|_{1}^{2}=\frac{{{e}^{2}}}{2}-e=e\left( \frac{e}{2}-1 \right)\]
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