A) \[\frac{1}{2}-e-\frac{1}{{{e}^{2}}}\]
B) \[-\frac{1}{2}+\frac{1}{e}-\frac{1}{2{{e}^{2}}}\]
C) \[\frac{3}{2}-e-\frac{1}{2{{e}^{2}}}\]
D) \[\frac{3}{2}-\frac{1}{e}-\frac{1}{2{{e}^{2}}}\]
Correct Answer: C
Solution :
Let\[I=\int\limits_{1}^{e}{\left\{ {{\left( \frac{x}{e} \right)}^{2x}}-{{\left( \frac{e}{x} \right)}^{x}} \right\}}{{\log }_{e}}xdx\] \[={{\int\limits_{1}^{e}{\left( \frac{x}{e} \right)}}^{2x}}{{\log }_{e}}xdx-\int\limits_{1}^{e}{{{\left( \frac{e}{x} \right)}^{x}}}{{\log }_{e}}xdx\] Let\[{{\left( \frac{x}{e} \right)}^{2x}}=u\]and\[{{\left( \frac{e}{x} \right)}^{x}}=v\] When\[x=1,u={{\left( \frac{1}{x} \right)}^{2}}\]and \[v=e\] And when\[x=e,u=1\]and\[v=1\] Now,\[2x\log \left( \frac{x}{e} \right)=\log u\] \[\Rightarrow \]\[2\left[ \log \left( \frac{x}{e} \right)+x\frac{e}{x}.\frac{1}{e} \right]dx=\frac{du}{u}\] \[\Rightarrow \]\[2[logx-1+1]dx=\frac{du}{u}\] \[\Rightarrow \]\[\log xdx=\frac{1}{2}.\frac{du}{u}\] and\[x\log \left( \frac{e}{x} \right)=\log v\Rightarrow x(1-logx)=logv\] \[\Rightarrow \]\[x-x\log x=\log v\] \[\Rightarrow \]\[\left( 1-\log x-x.\frac{1}{x} \right)dx=\frac{dv}{v}\] \[\Rightarrow \]\[-\log xdx=\frac{dv}{v}\] \[\therefore \]\[I=\frac{1}{2}\int\limits_{{{\left( \frac{1}{e} \right)}^{2}}}^{1}{u\frac{du}{u}}+\int\limits_{e}^{1}{v\frac{dv}{v}}\] \[=\frac{1}{2}\left( 1-\frac{1}{{{e}^{2}}} \right)+(1-e)=\frac{3}{2}-e-\frac{1}{2{{e}^{2}}}\]
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