Question

If the value of $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}=a$then find the value of $\int\limits_{e}^{{{e}^{4}}}{\sqrt{\ln x}dx}$
(a) $2{{e}^{4}}-e-a$
(b) ${{e}^{4}}-a$
(c) $2{{e}^{4}}-a$
(d) ${{e}^{4}}-e$

Answer 1

Hint: First, before proceeding for this, we must solve the given integral which is $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}=a$by using the substitution method to get the value of a. Then, by using the substitution for the integral $\int\limits_{e}^{{{e}^{4}}}{\sqrt{\ln x}dx}$ as $\ln x=z$and then by solving and differentiating both sides, we get the value of integral. Then, by using the integration by parts rule in which we have$\int{u\times vdx=}u\int{vdx}-\int{\dfrac{d}{dx}u}\times vdx$and by substituting the value of a, we get the value of integral which is required.

Complete step by step answer:
In this question, we are supposed to find the value of $\int\limits_{e}^{{{e}^{4}}}{\sqrt{\ln x}dx}$when the value of $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}=a$.
So, before proceeding for this, we must solve the given integral which is $\int\limits_{1}^{2}{{{e}^{{{x}^{2}}}}dx}=a$by using the substitution method as:
Suppose ${{x}^{2}}=t$ and then differentiating both sides, we get:
$\begin{align}
  & 2xdx=dt \\
 & \Rightarrow dx=\dfrac{dt}{2\sqrt{t}} \\
\end{align}$
Then, by substituting the value of dx and ${{x}^{2}}$ in the above integral, we get:
$\int\limits_{1}^{4}{\dfrac{{{e}^{t}}}{2\sqrt{t}}dt}=a$.....(i)
Now, by using the substitution for the integral $\int\limits_{e}^{{{e}^{4}}}{\sqrt{\ln x}dx}$ as $\ln x=z$and then by solving and differentiating both sides, we get:
$\begin{align}
  & x={{e}^{z}} \\
 & dx={{e}^{z}}dz \\
\end{align}$
Then, by substituting the value of dx and $\ln x=z$, we get the integral as:
$\int\limits_{1}^{4}{\sqrt{z}\centerdot {{e}^{z}}dz}$
Now, by using the integration by parts rule in which we have:
$\int{u\times vdx=}u\int{vdx}-\int{\dfrac{d}{dx}u}\times vdx$
Now, by using the above formula, we get:
$\begin{align}
  & \int\limits_{1}^{4}{\sqrt{z}\centerdot {{e}^{z}}dz}=\sqrt{z}\int\limits_{1}^{4}{{{e}^{z}}dz}-\int\limits_{1}^{4}{\dfrac{d}{dz}\sqrt{z}\times {{e}^{z}}dz} \\
 & \Rightarrow \int\limits_{1}^{4}{\sqrt{z}\centerdot {{e}^{z}}dz}=\left[ \sqrt{z}{{e}^{z}} \right]_{1}^{4}-\int\limits_{1}^{4}{\dfrac{1}{2\sqrt{z}}\times {{e}^{z}}dz} \\
\end{align}$
Now, b y using equation (i), we get the value of $\int\limits_{1}^{4}{\dfrac{1}{2\sqrt{z}}\times {{e}^{z}}dz}$as a, we get the above integral as:
$\begin{align}
  & \int\limits_{1}^{4}{\sqrt{z}\centerdot {{e}^{z}}dz}=\left[ \sqrt{4}{{e}^{4}}-\sqrt{1}e \right]-a \\
 & \Rightarrow \int\limits_{1}^{4}{\sqrt{z}\centerdot {{e}^{z}}dz}=2{{e}^{4}}-e-a \\
\end{align}$
So, we get the value of integral as $2{{e}^{4}}-e-a$.
Hence, option (a) is correct.

Note:
 Now, to solve these type of the questions we need to know some of the basics of differentiation so to get the answers correctly. So, the required formulas for the above question are:
$\begin{align}
  & \dfrac{d}{dx}{{e}^{x}}={{e}^{x}} \\
 & \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}} \\
\end{align}$