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Question: If we have the integral \(\int{{{x}^{5}}{{e}^{-{{x}^{2}}}}dx}=g\left( x \right){{e}^{-{{x}^{2}}}}+C\...

If we have the integral ∫x5e−x2dx=g(x)e−x2+C\int{{{x}^{5}}{{e}^{-{{x}^{2}}}}dx}=g\left( x \right){{e}^{-{{x}^{2}}}}+C, then the value of g(-1) is
[a] −52\dfrac{-5}{2}
[b] 1
[c] 12\dfrac{1}{2}
[d] -1

Explanation

Solution

Put −x2=t-{{x}^{2}}=t and hence prove that ∫e−x2x5dx=−12∫t2etdt\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−x2g(x)+C{{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=∫x5e−x2dxI=\int{{{x}^{5}}{{e}^{-{{x}^{2}}}}dx}
Put −x2=t-{{x}^{2}}=t
Differentiating with respect to x, we get
−2xdx=dt ⇒xdx=−dt2 \begin{aligned} & -2xdx=dt \\\ & \Rightarrow xdx=\dfrac{-dt}{2} \\\ \end{aligned}
Hence, we have
I=∫x5e−x2dx=∫e−x2(−x2)2xdx =∫ett2(−dt2)=−12∫t2etdt \begin{aligned} & 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{aligned}
We know that if ∫f(x)dx=u(x)\int{f\left( x \right)dx}=u\left( x \right) and ddx(g(x))=v(x)\dfrac{d}{dx}\left( g\left( x \right) \right)=v\left( x \right), then
∫f(x)g(x)dx=g(x)u(x)−∫v(x)u(x)dx\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(t)=etf\left( t \right)={{e}^{t}} and g(t)=t2g\left( t \right)={{t}^{2}}.
We have
∫f(t)dt=∫etdt=et\int{f\left( t \right)dt}=\int{{{e}^{t}}dt}={{e}^{t}} and ddt(g(t))=2t\dfrac{d}{dt}\left( g\left( t \right) \right)=2t
Hence, we have
I=−12(t2et−2∫tetdt)I=\dfrac{-1}{2}\left( {{t}^{2}}{{e}^{t}}-2\int{t{{e}^{t}}dt} \right)
Again applying integration by parts taking f(x)=etf\left( x \right)={{e}^{t}} and g(x)=tg\left( x \right)=t, we get
I=−t2et2+tet−∫etdt=−t2et2+tet−et+C =et(−t22+t−1)+C \begin{aligned} & 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{aligned}
Reverting to original variable, we get
I=e−x2(−x42−x2−1)+CI={{e}^{-{{x}^{2}}}}\left( \dfrac{-{{x}^{4}}}{2}-{{x}^{2}}-1 \right)+C
Hence, we have
g(x)e−x2=e−x2(−x42−x2−1) ⇒g(x)=(−x42−x2−1) \begin{aligned} & 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{aligned}
Hence, we have
g(−1)=(−12−1−1)=−52g\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 ∫t2etdt\int{{{t}^{2}}{{e}^{t}}dt} using the fact that ∫ex(f(x)+f′(x))=exf(x)+C\int{{{e}^{x}}\left( f\left( x \right)+f'\left( x \right) \right)}={{e}^{x}}f\left( x \right)+C
We have
∫t2etdt=∫et(t2+2t−2t−2+2)dt =∫et(t2+2t)dt+∫et(−2t−2)dt+2∫etdt =t2et−2tet+2et+C \begin{aligned} & \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{aligned}
Which is the same as obtained above.