JEE Main 12 Jan 2019 Evening Question 19
Question: The integral $ \int\limits_1^{e}{{ {{( \frac{x}{e} )}^{2x}}-{{( \frac{e}{x} )}^{x}} }}{\log_e}xdx $ equal to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ \frac{1}{2}-e-\frac{1}{e^{2}} $
B) $ -\frac{1}{2}+\frac{1}{e}-\frac{1}{2e^{2}} $
C) $ \frac{3}{2}-e-\frac{1}{2e^{2}} $
D) $ \frac{3}{2}-\frac{1}{e}-\frac{1}{2e^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ I=\int\limits_1^{e}{{ {{( \frac{x}{e} )}^{2x}}-{{( \frac{e}{x} )}^{x}} }}{\log_e}xdx $ $ ={{\int\limits_1^{e}{( \frac{x}{e} )}}^{2x}}{\log_e}xdx-\int\limits_1^{e}{{{( \frac{e}{x} )}^{x}}}{\log_e}xdx $
Let $ {{( \frac{x}{e} )}^{2x}}=u $ and $ {{( \frac{e}{x} )}^{x}}=v $
When $ x=1,u={{( \frac{1}{x} )}^{2}} $ and $ v=e $
And when $ x=e,u=1 $ and $ v=1 $ Now, $ 2x\log ( \frac{x}{e} )=\log u $
$ \Rightarrow $ $ 2[ \log ( \frac{x}{e} )+x\frac{e}{x}.\frac{1}{e} ]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 ( \frac{e}{x} )=\log v\Rightarrow x(1-logx)=logv $
$ \Rightarrow $ $ x-x\log x=\log v $
$ \Rightarrow $ $ ( 1-\log x-x.\frac{1}{x} )dx=\frac{dv}{v} $
$ \Rightarrow $ $ -\log xdx=\frac{dv}{v} $
$ \therefore $ $ I=\frac{1}{2}\int\limits _{{{( \frac{1}{e} )}^{2}}}^{1}{u\frac{du}{u}}+\int\limits_e^{1}{v\frac{dv}{v}} $ $ =\frac{1}{2}( 1-\frac{1}{e^{2}} )+(1-e)=\frac{3}{2}-e-\frac{1}{2e^{2}} $
