Science:Math Exam Resources/Courses/MATH101/April 2018/Question 04 (i)

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MATH101 April 2018

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Question 04 (i)
Write an integral that corresponds to the limit as ${\textstyle n\to \infty }$ of the Riemann sum $\sum _{i=1}^{n}{\frac {4i}{n^{2}}}e^{-2i/n}\,.$ Then calculate the resulting integral explicitly.

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
Start with $x_{i}={\frac {i}{n}}$ .

Hint 2
Consider the integration by parts to evaluate the integral.

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.
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Solution
As suggested in the Hint, start with $x_{i}={\frac {i}{n}}$ and $\Delta x=x_{i+1}-x_{i}={\frac {1}{n}}$ . Then, the Riemann sum can be written as $\sum _{i=1}^{n}{\frac {4i}{n^{2}}}e^{-2i/n}=\sum _{i=1}^{n}4x_{i}e^{-2x_{i}}\Delta x.$ Observe that $x_{1}={\frac {1}{n}}\to 0$ as $n\to \infty$ and $x_{n}=1$ for any $n\in \mathbb {N}$ . Based on this, we put $f(x)=4xe^{-2x}$ to get $\sum _{i=1}^{n}4x_{i}e^{-2x_{i}}\Delta x=\sum _{i=1}^{n}f(x_{i})\Delta x\to \int _{0}^{1}f(x)dx=\int _{0}^{1}4xe^{-2x}dx,$ as $n\to \infty$ . Now, we evaluate the integral. Applying the integration by parts with $f(x)=4x$ , $g(x)=e^{-2x}$ , $f'(x)=4$ , and $g'(x)=-2e^{-2x}$ , we have ${\begin{aligned}\int _{0}^{1}4xe^{-2x}dx&=-{\frac {1}{2}}\int _{0}^{1}f(x)g'(x)dx=-{\frac {1}{2}}\left[{\bigg (}f(x)g(x){\bigg \|}_{0}^{1}-\int _{0}^{1}f'(x)g(x)dx\right]\\&=-{\frac {1}{2}}(4e^{-2}-0)+{\frac {1}{2}}\int _{0}^{1}4e^{-2x}dx=-2e^{-2}+2\int _{0}^{1}e^{-2x}dx\\&=-2e^{-2}+2{\bigg (}-{\frac {1}{2}}e^{-2x}{\bigg \|}_{0}^{1}\\&=-2e^{-2}+2\cdot (-{\frac {1}{2}})\cdot {\bigg (}e^{-2x}{\bigg \|}_{0}^{1}\\&=-2e^{-2}-(e^{-2}-e^{0})=-3e^{-2}+1.\end{aligned}}$ Answer: $\color {blue}-3e^{-2}+1$