Question

Given $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx=a}$ , then the value of $\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}$ is.
(a) ${{e}^{4}}-e$
(b) ${{e}^{4}}-a$
(c) $2{{e}^{4}}-a$
(d) $2{{e}^{4}}-e-a$

Answer 1

Hint: For solving this question we will use integration by parts formula to do the integration and then we will try to evaluate the given integral correctly.

Complete step-by-step solution -
Given:
It is given that, $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx=a}$ and we have to find the value of $\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}$ .
Now, let $I=\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}$ and $t=\sqrt{{{\log }_{e}}x}$ . Then,
$\begin{align}
  & t=\sqrt{{{\log }_{e}}x} \\
 & {{t}^{2}}={{\log }_{e}}x \\
 & {{e}^{{{t}^{2}}}}=x \\
 & 2t{{e}^{{{t}^{2}}}}dt=dx \\
\end{align}$
And, when $x=e$ then $t=1$ and when $x={{e}^{4}}$ then $t=2$ . Then,
\[\begin{align}
  & I=\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}=\int\limits_{1}^{2}{2{{t}^{2}}{{e}^{{{t}^{2}}}}dt} \\
 & \Rightarrow I=\int\limits_{1}^{2}{2{{t}^{2}}{{e}^{{{t}^{2}}}}dt} \\
\end{align}\]
Now, in the above expression of $I$ as $t$ is just a variable so we can write, \[\begin{align}
  & I=\int\limits_{1}^{2}{2{{t}^{2}}{{e}^{{{t}^{2}}}}dt}=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx} \\
 & \Rightarrow I=\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx} \\
\end{align}\]
Now, before we proceed to solve the above integral we should know the formula of integration by parts which are written below:
$\int\limits_{l}^{u}{f\left( x \right)}\cdot g\left( x \right)dx=\left[ f\left( x \right)\int{g\left( x \right)}dx \right]_{l}^{u}-\int\limits_{l}^{u}{\left[ \left( \dfrac{d\left( f\left( x \right) \right)}{dx} \right)\cdot \left( \int{g\left( x \right)dx} \right) \right]dx}$
Now, we will use the above formula directly to evaluate the value of integral \[\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}\] . Where $f\left( x \right)={{e}^{{{x}^{2}}}}$ , $g\left( x \right)=1$ , $l=1$ and $u=2$ . Then,
\[\begin{align}
  & \int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}=a=\left[ {{e}^{{{x}^{2}}}}\int{dx} \right]_{1}^{2}-\int\limits_{1}^{2}{\left[ \left( \dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx} \right)\left( \int{dx} \right) \right]dx} \\
 & \Rightarrow a=\left[ {{e}^{{{x}^{2}}}}\cdot x \right]_{1}^{2}-\int\limits_{1}^{2}{\left[ \left( 2x{{e}^{{{x}^{2}}}} \right)x \right]}dx \\
 & \Rightarrow a=\left[ 2{{e}^{4}}-e \right]-\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx} \\
 & \Rightarrow a=\left[ 2{{e}^{4}}-e \right]-\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx} \\
 & \Rightarrow \int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx}=2{{e}^{4}}-e-a \\
\end{align}\]
Now, from the above equation, we will find that \[\int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx}=I\] . Then, we conclude that:
\[\begin{align}
  & \int\limits_{1}^{2}{2{{x}^{2}}{{e}^{{{x}^{2}}}}dx}=2{{e}^{4}}-e-a \\
 & \Rightarrow I=2{{e}^{4}}-e-a \\
\end{align}\]
Now, as per our assumption $I=\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}$ . Then,
$\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}=2{{e}^{4}}-e-a$
Thus, the value of $\int\limits_{e}^{{{e}^{4}}}{\sqrt{{{\log }_{e}}x}dx}$ is $2{{e}^{4}}-e-a$ .
Hence, option (d) is the correct option.

Note: Here, the student should do the correct substitution and for any such problem we should proceed stepwise while solving without doing any calculation mistake. Similarly, before exactly solving the integral first we should analyse what we have to evaluate and how to start the solution, so that we can get the value of the given integral. Then, select the correct option without any error.