Question

The integral \int\limits_1^e {\left\\{ {{{\left( {\dfrac{x}{e}} \right)}^{2x}} - {{\left( {\dfrac{e}{x}} \right)}^x}} \right\\}{{\log }_e}xdx} 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

Hint : In order to solve the question given above, you need to have a basic understanding of the concepts of integration. Since, it is visible that no direct formula can be applied to solve this question, we have to equate the value of ${\left( {\dfrac{x}{e}} \right)^x}$ in the equation. This will help in simplifying the integration of the above integral.

Complete step by step solution:
We are given \int\limits_1^e {\left\\{ {{{\left( {\dfrac{x}{e}} \right)}^{2x}} - {{\left( {\dfrac{e}{x}} \right)}^x}} \right\\}{{\log }_e}xdx}
To solve the above question,
We have to put

{\left( {\dfrac{x}{e}} \right)^x} = t \\\ \Rightarrow x{\log _e}\left( {\dfrac{x}{e}} \right) = \log t \; $$ . Now, we will use the subtraction principle of logarithm.

x\left( {{{\log }_e}x - {{\log }_e}e} \right) = \log t \\
\Rightarrow \left[ {x\left( {\dfrac{1}{x}} \right) + \left( {{{\log }_e}x - {{\log }_e}e} \right)} \right] dx = \dfrac{1}{t}dt \\

$$On further solving the equation, we get:$$

\left( {1 + {{\log }_e}x - 1} \right)dx = \dfrac{1}{t}dt \\
\Rightarrow \left( {{{\log }_e}x} \right)dx = \dfrac{1}{t}dt ;

$$Now, we have to find the limits, Upper limit:$$

x = e \\
\Rightarrow t = 1 \\

$$And lower limit:$$

x = 1 \\
\Rightarrow t = \dfrac{1}{e} ;

$$Putting these values in the above integral, it changes to:$$

\int\limits_{\dfrac{1}{e}}^1 {\left( {{t^2} - \dfrac{1}{t}} \right)} \times \dfrac{1}{t}dt \\
\Rightarrow \int\limits_{\dfrac{1}{e}}^1 {\left( {t - \dfrac{1}{{{t^2}}}} \right)} dt ;

Now, integrate the above integral. After integration, we will get: $${\left[ {\left( {\dfrac{{{t^2}}}{2} + \dfrac{1}{t}} \right)} \right] ^1}_{\dfrac{1}{e}}$$ On equating the upper and the lower limits, we will get, $$\left( {\dfrac{1}{2} + 1} \right) - \left( {\dfrac{1}{{2{e^2}}} + e} \right)$$ $$ \Rightarrow \dfrac{3}{2} - \dfrac{1}{{2{e^2}}} - e$$ . This can also be written as $$\dfrac{3}{2} - e - \dfrac{1}{{2{e^2}}}$$ . **So, the correct answer is “Option B”.** **Note** : While solving questions similar to the one given above, always remember the concepts of integration. You need to know that whenever you cannot solve a question through application of a formula or direct integration is complex, you can substitute values in the integral to simplify it. In the above integration we have used the method of substitution to make the integration of the integral quite simple. Also, remember whenever you have an integral with limits (or a definite integral) remember to equate the limits after integration.