A) \[e\sqrt{e}+1\]
B) \[e\sqrt{e}-2\]
C) \[2(e\sqrt{e}-1)\]
D) \[e\sqrt{e}-1\]
Correct Answer: D
Solution :
[d] Let \[I=\int\limits_{\frac{1}{2}}^{2}{{{e}^{\left| x-\frac{1}{x} \right|}}}\,\,dx\] ?..(1) Putting \[x=\frac{1}{t},\] we get \[\therefore \,\,\,I=-\int\limits_{2}^{\frac{1}{2}}{{{e}^{\left| t-\frac{1}{t} \right|}}}\frac{dt}{{{t}^{2}}}\] ? (2) Adding (1) and (2), we get \[2I=\int\limits_{\frac{1}{2}}^{2}{{{e}^{\left| x-\frac{1}{x} \right|}}.\left( 1+\frac{1}{{{x}^{2}}} \right)}dx\] \[\therefore \,\,\,2I=\int\limits_{\frac{1}{2}}^{2}{{{e}^{-\left( x-\frac{1}{x} \right)}}.\left( 1+\frac{1}{{{x}^{2}}} \right)}dx+\int\limits_{1}^{2}{{{e}^{\left( x-\frac{1}{x} \right)}}}.\left( 1+\frac{1}{{{x}^{2}}} \right)dx\]\[=\int\limits_{-\frac{3}{2}}^{0}{{{e}^{-y}}}dy+\int\limits_{0}^{\frac{3}{2}}{{{e}^{y}}\,dy}\] \[\therefore \,\,\,2I=2\left( e\sqrt{e}-1 \right)\] \[\Rightarrow \,\,I=e\sqrt{e}-1\]
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