Science:Math Exam Resources/Courses/MATH101/April 2017/Question 09

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MATH101 April 2017

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Question 09
Evaluate $\int _{0}^{\infty }6x^{3}e^{-x^{2}}dx$ . Simplify your answer completely.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Simplify the integral with the substitution $u=x^{2}$ . Then use the integration by parts on the resulting integral.

Hint 2
(For the alternative solution) Rewrite $6x^{3}e^{-x^{2}}$ as a product so that one of the factors is the derivative of $e^{-x^{2}}$ and use integration by parts.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1
Using the substitution $u=x^{2}$ , we have $du=2x\,dx$ , so that $\int 6x^{3}e^{-x^{2}}\,dx=\int 3x^{2}e^{-x^{2}}\cdot 2x\,dx=\int 3ue^{-u}\,du$ Then we apply the integration by parts for $f(u)=u$ and $g(u)=-e^{-u}$ to get $\int ue^{-u}\,du=-ue^{-u}-\int -e^{-u}\,du=-ue^{-u}-e^{-u}+C$ Plugging $u=x^{2}$ back, $\int 6x^{3}e^{-x^{2}}\,dx=3(-ue^{-u}-e^{-u}+C)=-3x^{2}e^{-x^{2}}-3e^{-x^{2}}+C$ and so $\int _{0}^{\infty }6x^{3}e^{-x^{2}}\,dx=\lim _{t\to \infty }\int _{0}^{t}6x^{3}e^{-x^{2}}\,dx=\lim _{t\to \infty }-3x^{2}e^{-x^{2}}-3e^{-x^{2}}{\bigg \|}_{x=0}^{t}=\lim _{t\to \infty }(-3t^{2}e^{-t^{2}}-3e^{-t^{2}}+3)=3$ In the last equality, we use the L'hospital's rule to evaluate ${\begin{aligned}\lim _{t\to \infty }t^{2}e^{-t^{2}}=\lim _{t\to \infty }{\frac {t^{2}}{e^{t^{2}}}}=\lim _{t\to \infty }{\frac {(t^{2})'}{(e^{t^{2}})'}}=\lim _{t\to \infty }{\frac {2t}{2te^{t^{2}}}}=\lim _{t\to \infty }{\frac {1}{e^{t^{2}}}}=0.\end{aligned}}$ Therefore, we get $\int _{0}^{\infty }6x^{3}e^{-x^{2}}=\color {blue}3$

Solution 2
(Alternative solution) Let us rewrite the integrand, so that $\int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=-3\int _{0}^{\infty }x^{2}(-2xe^{-x^{2}})dx$ Use integration by parts where $f^{\prime }=-2xe^{-x^{2}},\,g=x^{2}\,,f=e^{-x^{2}},\,g'=2x$ and so $\int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=-3[x^{2}e^{-x^{2}}\|_{0}^{\infty }-\int _{0}^{\infty }2xe^{-x^{2}}dx$ $=-3[x^{2}e^{-x^{2}}\|_{0}^{\infty }+e^{-x^{2}}\|_{0}^{\infty }\,]$ Note that $\lim _{x\rightarrow \infty }x^{2}e^{-x^{2}}=0$ and $\lim _{x\rightarrow \infty }e^{-x^{2}}=0$ Thus, from the first term we have $-3(0-0)=0$ and from the 2nd term we get $-3(0-1)=3$ Answer: $\color {blue}\int _{0}^{\infty }6x^{3}e^{-x^{2}}dx=3$