Question

The value of integral \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}\] is
(a) \[\dfrac{6!}{{{2}^{7}}}\]
(b) \[\dfrac{6!}{{{2}^{6}}}\]
(c) \[{{2}^{6}}6!\]
(d) \[{{2}^{7}}6!\]

Answer 1

Hint: Integrate the given function using product rule of integration of two functions. Apply the limits to evaluate the value of function near infinity and zero. Substitute the values to calculate the definite integral of the given function.

Complete step by step answer:
We have to calculate the value of the integral \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}\]. We will use the product rule of integration of two functions to get the integration of a given function and then apply the limits.
We know that integration of product of two functions \[u\left( x \right)v\left( x \right)\] is given by \[\int{u\left( x \right)v\left( x \right)dx=u\left( x \right)\int{v\left( x \right)dx-\int{\left( \dfrac{d}{dx}u\left( x \right)\int{v\left( x \right)dx} \right)dx}}}\].
Substituting \[u\left( x \right)={{e}^{\dfrac{-x}{2}}},v\left( x \right)={{x}^{6}}\] in the above equation, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}={{x}^{6}}\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d}{dx}\left( {{x}^{6}} \right)\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx}.....\left( 1 \right)\].
We know that integration of \[{{e}^{ax}}\] is \[\int{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a}\].
Substituting \[a=\dfrac{-1}{2}\] in the above equation, we have \[\int{{{e}^{-\dfrac{x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}.....\left( 2 \right)\].
We know that differentiation of \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=6,b=0\] in the above equation, we have \[\dfrac{d}{dx}\left( {{x}^{6}} \right)=6{{x}^{5}}.....\left( 3 \right)\].
Substituting equation (2) and (3) in equation (1), we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}={{x}^{6}}\left( -2{{e}^{\dfrac{-x}{2}}} \right)-\int{6{{x}^{5}}\left( -2{{e}^{\dfrac{-x}{2}}} \right)dx}\].
We can rewrite the above equation as \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}+12\int{{{e}^{\dfrac{-x}{2}}}{{x}^{5}}dx}\] because \[\int{cf\left( x \right)dx}=c\int{f\left( x \right)dx}\], where c is a constant.
We will further simplify the above equation using the product rule of integration of two functions.
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}+12\int{{{e}^{\dfrac{-x}{2}}}{{x}^{5}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}+12\left( {{x}^{5}}\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d\left( {{x}^{5}} \right)}{dx}\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx} \right)\].
Simplifying the above expression, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}+12\left( {{x}^{5}}\left( -2{{e}^{\dfrac{-x}{2}}} \right)-\int{-2{{e}^{\dfrac{-x}{2}}}\left( 5{{x}^{4}} \right)dx} \right)\].
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}+120\int{{{e}^{\dfrac{-x}{2}}}{{x}^{4}}dx}\].
Further simplifying the above expression, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}+120\left( {{x}^{4}}\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d\left( {{x}^{4}} \right)}{dx}\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx} \right)\].
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}+960\int{{{x}^{3}}{{e}^{\dfrac{-x}{2}}}dx}\].
Simplifying the above expression, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}+960\left( {{x}^{3}}\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d\left( {{x}^{3}} \right)}{dx}\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx} \right)\].
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}+5760\int{{{x}^{2}}{{e}^{\dfrac{-x}{2}}}dx}\].
Simplifying the above expression, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}+5760\left( {{x}^{2}}\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d\left( {{x}^{2}} \right)}{dx}\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx} \right)\].
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}+23040\int{x{{e}^{\dfrac{-x}{2}}}dx}\].
Simplifying the above expression, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}+23040\left( x\int{{{e}^{\dfrac{-x}{2}}}dx}-\int{\left( \dfrac{d\left( x \right)}{dx}\int{{{e}^{\dfrac{-x}{2}}}dx} \right)dx} \right)\].
Thus, we have \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}-46080x{{e}^{\dfrac{-x}{2}}}-92160{{e}^{\dfrac{-x}{2}}}\].
We have evaluated the value of the integral \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}\].
We will now evaluate the value of integral at \[x=0\] and \[x\to \infty \].
We will firstly evaluate the value of \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}-46080x{{e}^{\dfrac{-x}{2}}}-92160{{e}^{\dfrac{-x}{2}}}\] as \[x\to \infty \].
As \[x\to \infty \], we observe \[{{e}^{\dfrac{-x}{2}}}\to {{e}^{-\infty }}=0\].
Thus, we have \[{{\left[ -2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}-46080x{{e}^{\dfrac{-x}{2}}}-92160{{e}^{\dfrac{-x}{2}}} \right]}_{x\to \infty }}=0\].
Now, we will evaluate the value of \[\int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}=-2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}-46080x{{e}^{\dfrac{-x}{2}}}-92160{{e}^{\dfrac{-x}{2}}}\] at \[x=0\].
At \[x=0\], we have \[{{\left[ -2{{e}^{\dfrac{-x}{2}}}{{x}^{6}}-24{{e}^{\dfrac{-x}{2}}}{{x}^{5}}-240{{x}^{4}}{{e}^{\dfrac{-x}{2}}}-1920{{x}^{3}}{{e}^{\dfrac{-x}{2}}}-11520{{x}^{2}}{{e}^{\dfrac{-x}{2}}}-46080x{{e}^{\dfrac{-x}{2}}}-92160{{e}^{\dfrac{-x}{2}}} \right]}_{x=0}}=-92160\].
We know that \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}={{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x\to \infty }}-{{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x=0}}\].
Thus, we have \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}={{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x\to \infty }}-{{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x=0}}=0-\left( -92160 \right)=92160\].
We can rewrite 92160 as \[92160={{2}^{7}}\times 720={{2}^{7}}\times 6!\].
Thus, we have \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}={{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x\to \infty }}-{{\left[ \int{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx} \right]}_{x=0}}=0-\left( -92160 \right)=92160=92160={{2}^{7}}\times 6!\].
Hence, the value of integral \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}\] is \[{{2}^{7}}\times 6!\].

So, Option (d) is the right answer.

Note: The integration of \[\int\limits_{0}^{\infty }{{{x}^{6}}{{e}^{\dfrac{-x}{2}}}dx}\]represents the area of the function ${{x}^{6}}{{e}^{\dfrac{-x}{2}}}$ in the range $\left( 0,\infty \right)$. One must be careful while applying the limit \[x\to \infty \] to the integral. If we simply substitute $x=\infty $, we will get an incorrect answer. It’s necessary to use the formula for the integral product of two functions. Otherwise, we won’t be able to solve this question.