Question

The value of the integral \int\limits_{1}^{e}{\left\\{ {{\left( \dfrac{x}{e} \right)}^{2x}}-{{\left( \dfrac{e}{x} \right)}^{x}} \right\\}\ln x\ dx} is equal to:
A. $\dfrac{1}{2}-e-\dfrac{1}{{{e}^{2}}}$
B. $\dfrac{3}{2}-\dfrac{1}{e}-\dfrac{1}{2{{e}^{2}}}$
C. $-\dfrac{1}{2}+\dfrac{1}{e}-\dfrac{1}{2{{e}^{2}}}$
D. $\dfrac{3}{2}-e-\dfrac{1}{2{{e}^{2}}}$

Answer 1

To solve this question, we should use the method of substitutions in integration. Let us substitute ${{\left( \dfrac{x}{e} \right)}^{x}}=t$ and apply logarithm on both sides, we get
$\begin{aligned} & \ln \left( {{\left( \dfrac{x}{e} \right)}^{x}} \right)=\ln t \\\ & \Rightarrow x\ln \left( \dfrac{x}{e} \right)=\ln t \\\ \end{aligned}$
Differentiating on both sides, we get
$\begin{aligned} & \left( \ln \left( \dfrac{x}{e} \right)+x\times \dfrac{e}{x}\times \dfrac{1}{e} \right)dx=\dfrac{1}{t}dt \\\ & \Rightarrow \left( \ln x-\ln e+1 \right)dx=\dfrac{1}{t}dt \\\ & \Rightarrow \ln xdx=\dfrac{1}{t}dt \\\ \end{aligned}$
Using these substitutions in the integral with the updated limits, we get a simplified form of integration and we can get the answer.

Complete step-by-step answer:
We are asked the value of the integral \int\limits_{1}^{e}{\left\\{ {{\left( \dfrac{x}{e} \right)}^{2x}}-{{\left( \dfrac{e}{x} \right)}^{x}} \right\\}\ln x\ dx}.
To get the answer, we should use a method of substitutions. Let us consider the substitution
${{\left( \dfrac{x}{e} \right)}^{x}}=t\to \left( 1 \right)$.
Let us apply the logarithm base e to the above equation, we get
$\ln \left( {{\left( \dfrac{x}{e} \right)}^{x}} \right)=\ln t$
We know the formula $\ln {{x}^{a}}=a\ln x$. Using this in the above equation, we get
$x\ln \left( \dfrac{x}{e} \right)=\ln t$
Differentiating on both sides, we get
$d\left( x\ln \left( \dfrac{x}{e} \right) \right)=d\left( \ln t \right)$
We know the product rule in integration which is
$d\left( uv \right)=ud\left( v \right)+vd\left( u \right)$

Using this in the above equation, we get
$xd\left( \ln \left( \dfrac{x}{e} \right) \right)+\ln \left( \dfrac{x}{e} \right)d\left( x \right)=d\left( \ln t \right)$
We know that $d\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)$ which is called the chain rule and $d\left( \ln x \right)=\dfrac{1}{x}dx$
Using them in the above equation, we get
$\left( x\times \dfrac{1}{\dfrac{x}{e}}\times d\left( \dfrac{x}{e} \right)+\ln \left( \dfrac{x}{e} \right)\times 1 \right)dx=\dfrac{1}{t}dt$
We can simplify it as
$\begin{aligned} & \left( x\times \dfrac{e}{x}\times \dfrac{1}{e}+\ln \left( \dfrac{x}{e} \right) \right)dx=\dfrac{1}{t}dt \\\ & \left( 1+\ln \left( \dfrac{x}{e} \right) \right)dx=\dfrac{1}{t}dt \\\ \end{aligned}$
We know that $\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b$
Using it in above equation, we get
$\left( 1+\ln x-\ln e \right)dx=\dfrac{1}{t}dt$
We know that $\ln e=1$. Using it, we get
$\begin{aligned} & \left( 1+\ln x-1 \right)dx=\dfrac{1}{t}dt \\\ & \Rightarrow \ln x\ dx=\dfrac{1}{t}dt\to \left( 2 \right) \\\ \end{aligned}$
Using the equation-1 , the limits of integration will be
$\begin{aligned} & x=e\Rightarrow {{\left( \dfrac{e}{e} \right)}^{e}}=t \\\ & \Rightarrow t=1 \\\ \end{aligned}$
$\begin{aligned} & x=1\Rightarrow {{\left( \dfrac{1}{e} \right)}^{1}}=t \\\ & \Rightarrow t=\dfrac{1}{e} \\\ \end{aligned}$
The integral can be modified in terms of t as
I=\int\limits_{1}^{e}{\left\\{ {{\left( \dfrac{x}{e} \right)}^{2x}}-{{\left( \dfrac{e}{x} \right)}^{x}} \right\\}\ln x\ dx}\Rightarrow I=\int\limits_{\dfrac{1}{e}}^{1}{\left( {{t}^{2}}-\dfrac{1}{t} \right)\dfrac{1}{t}\ dt}
The integral becomes
$I=\int\limits_{\dfrac{1}{e}}^{1}{\left( t-\dfrac{1}{{{t}^{2}}} \right)\ dt}$
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$. Using it in above integral, we get
$I=\left[ \dfrac{{{t}^{1+1}}}{1+1}-\dfrac{{{t}^{-2+1}}}{-2+1} \right]_{\dfrac{1}{e}}^{1}=\left[ \dfrac{{{t}^{2}}}{2}+\dfrac{1}{t} \right]_{\dfrac{1}{e}}^{1}$
By substituting the limits, we get
$I=\left[ \left( \dfrac{{{1}^{2}}}{2}+\dfrac{1}{1} \right)-\left( \dfrac{{{\left( \dfrac{1}{e} \right)}^{2}}}{2}+\dfrac{1}{\dfrac{1}{e}} \right) \right]=\left[ \dfrac{3}{2}-\dfrac{1}{2{{e}^{2}}}-e \right]$
So, the value of the integral is
$I=\left[ \dfrac{3}{2}-e-\dfrac{1}{2{{e}^{2}}} \right]$
$\therefore$The value of the given integral is $I=\left[ \dfrac{3}{2}-e-\dfrac{1}{2{{e}^{2}}} \right]$.

So, the correct answer is “Option D”.

Note: The main trick in the question is the substitution that we did in the process. If we substitute another function, we may not get the result easily. For example, if we substitute $\ln x=t$, we can see that the integral will not be simplified. Students might make a mistake while differentiating the equation$x\ln \left( \dfrac{x}{e} \right)=\ln t$. Majority of the students forget to multiply the differentiation of $\dfrac{x}{e}$ while we differentiate $\ln \left( \dfrac{x}{e} \right)$. They do this in a hurry to get the answer. So, we should carefully differentiate where there is a function and be clear about the chain rule.