Question

If we have the integral $\int{{{x}^{5}}{{e}^{-{{x}^{2}}}}dx}=g\left( x \right){{e}^{-{{x}^{2}}}}+C$, then the value of g(-1) is
[a] $\dfrac{-5}{2}$
[b] 1
[c] $\dfrac{1}{2}$
[d] -1

Answer 1

Hint: Put $-{{x}^{2}}=t$ and hence prove that $\int{{{e}^{-{{x}^{2}}}}{{x}^{5}}dx}=-\dfrac{1}{2}\int{{{t}^{2}}{{e}^{t}}dt}$. Use integration by parts and hence evaluate the integral. Revert back to the original variable and equate the expression to ${{e}^{-{{x}^{2}}}}g\left( x \right)+C$. Compare like terms and hence find the value of g(x). Hence find the value of g(-1).

Complete step-by-step solution:
Let $I=\int{{{x}^{5}}{{e}^{-{{x}^{2}}}}dx}$
Put $-{{x}^{2}}=t$
Differentiating with respect to x, we get
$\begin{align}
  & -2xdx=dt \\
 & \Rightarrow xdx=\dfrac{-dt}{2} \\
\end{align}$
Hence, we have
$\begin{align}
  & I=\int{{{x}^{5}}}{{e}^{-{{x}^{2}}}}dx=\int{{{e}^{-{{x}^{2}}}}{{\left( -{{x}^{2}} \right)}^{2}}xdx} \\
 & =\int{{{e}^{t}}{{t}^{2}}\left( \dfrac{-dt}{2} \right)}=-\dfrac{1}{2}\int{{{t}^{2}}{{e}^{t}}dt} \\
\end{align}$
We know that if $\int{f\left( x \right)dx}=u\left( x \right)$ and $\dfrac{d}{dx}\left( g\left( x \right) \right)=v\left( x \right)$, then
$\int{f\left( x \right)g\left( x \right)dx}=g\left( x \right)u\left( x \right)-\int{v\left( x \right)u\left( x \right)dx}$. This is known as integration by parts rule. The function f(x) is called the second function and the function g(x) is called the first function.
Taking $f\left( t \right)={{e}^{t}}$ and $g\left( t \right)={{t}^{2}}$.
We have
$\int{f\left( t \right)dt}=\int{{{e}^{t}}dt}={{e}^{t}}$ and $\dfrac{d}{dt}\left( g\left( t \right) \right)=2t$
Hence, we have
$I=\dfrac{-1}{2}\left( {{t}^{2}}{{e}^{t}}-2\int{t{{e}^{t}}dt} \right)$
Again applying integration by parts taking $f\left( x \right)={{e}^{t}}$ and $g\left( x \right)=t$, we get
$\begin{align}
& I=-\dfrac{{{t}^{2}}{{e}^{t}}}{2}+t{{e}^{t}}-\int{{{e}^{t}}dt}=-\dfrac{{{t}^{2}}{{e}^{t}}}{2}+t{{e}^{t}}-{{e}^{t}}+C \\
 & ={{e}^{t}}\left( \dfrac{-{{t}^{2}}}{2}+t-1 \right)+C \\
\end{align}$
Reverting to original variable, we get
$I={{e}^{-{{x}^{2}}}}\left( \dfrac{-{{x}^{4}}}{2}-{{x}^{2}}-1 \right)+C$
Hence, we have
$\begin{align}
  & g\left( x \right){{e}^{-{{x}^{2}}}}={{e}^{-{{x}^{2}}}}\left( \dfrac{-{{x}^{4}}}{2}-{{x}^{2}}-1 \right) \\
 & \Rightarrow g\left( x \right)=\left( \dfrac{-{{x}^{4}}}{2}-{{x}^{2}}-1 \right) \\
\end{align}$
Hence, we have
$g\left( -1 \right)=\left( \dfrac{-1}{2}-1-1 \right)=\dfrac{-5}{2}$
Hence option [a] is correct.

Note: Alternative Solution:
We can simplify the integral $\int{{{t}^{2}}{{e}^{t}}dt}$ using the fact that $\int{{{e}^{x}}\left( f\left( x \right)+f'\left( x \right) \right)}={{e}^{x}}f\left( x \right)+C$
We have
$\begin{align}
  & \int{{{t}^{2}}{{e}^{t}}dt}=\int{{{e}^{t}}\left( {{t}^{2}}+2t-2t-2+2 \right)}dt \\
 & =\int{{{e}^{t}}\left( {{t}^{2}}+2t \right)dt}+\int{{{e}^{t}}\left( -2t-2 \right)dt}+2\int{{{e}^{t}}dt} \\
 & ={{t}^{2}}{{e}^{t}}-2t{{e}^{t}}+2{{e}^{t}}+C \\
\end{align}$
Which is the same as obtained above.