Science:Math Exam Resources/Courses/MATH103/April 2016/Question 01 (a)

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MATH103 April 2016

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Question 01 (a)
Express $\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{n}}e^{-k^{2}/n^{2}}$ as a definite integral of the form $\int _{0}^{1}f(x)\,dx.$ . Do not evaluate the integral.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that $\int _{a}^{b}f(x)\,dx=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f(x_{k})\Delta x$ , where $\Delta x={\frac {b-a}{n}}$ and $x_{k}=a+k\Delta x$ . Note: The sums above are called right Riemann sums, since the points at which the integrand is evaluated are the right endpoints of the subintervals $[x_{k},x_{k+1}],k=0,1,\dots ,n-1$ .

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Solution
We write the given series as a right Riemann sum, i.e., $\sum _{k=1}^{n}{\frac {1}{n}}e^{-{\frac {k^{2}}{n^{2}}}}=\sum _{k=1}^{n}f(x_{k})\Delta x$ , and try to find $\Delta x$ and $x_{k}$ , and in turn $f(x_{k})$ . The given integral has lower and upper bounds $a=0,b=1$ , respectively, so $\Delta x={\frac {b-a}{n}}={\frac {1}{n}}$ , which accounts for the first factor of each term in the series. Therefore, the second factor $e^{-{\frac {k^{2}}{n^{2}}}}$ must be $f(x_{k})$ . We know that $x_{k}=a+k\Delta x=0+k\Delta x=k\Delta x={\frac {k}{n}}$ , so we can write $e^{-{\frac {k^{2}}{n^{2}}}}$ in terms of $x_{k}$ as follows: ${\begin{aligned}e^{-{\frac {k^{2}}{n^{2}}}}&=e^{-{({\frac {k}{n}})}^{2}}\\&=e^{-x_{k}^{2}},\end{aligned}}$ which implies that $f(x_{k})=e^{-x_{k}^{2}}$ , whence $f(x)=e^{-x^{2}}$ . Therefore $\color {blue}\lim _{n\rightarrow \infty }\sum _{k=0}^{n}{\frac {1}{n}}e^{-{\frac {k^{2}}{n^{2}}}}=\int _{0}^{1}e^{-x^{2}}dx.$