Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (g)

MATH105 April 2012

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Question 08 (g)
Short-answer questions! No credit will be given for the answer (even if it is correct) without the accompanying work. For a certain function ƒ(x), the following equation holds: $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}{\frac {k}{n^{2}}}{\sqrt {1-{\frac {k^{2}}{n^{2}}}}}=\int _{0}^{1}f(x)\,dx.$ Find ƒ(x).

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 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
Alarm bells should be triggered into thinking this is a Riemann sum. Recall that for a function $f(x)$ we have that the general formula for a Riemann sum is $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\Delta xf(x_{k})=\int _{a}^{b}f(x)\,dx$ where $\Delta x={\frac {b-a}{n}}$ and lastly, $x_{k}=a+k\Delta x$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We wish to find $f(x)$ so that $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}{\frac {k}{n^{2}}}{\sqrt {1-{\frac {k^{2}}{n^{2}}}}}=\int _{0}^{1}f(x)dx.$ This requires making the limit of the Riemann sum (the expression on the left) look more like: $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\Delta xf(x_{k})$. Our interval is $[a,b]=[0,1]$ and this makes $\Delta x={\frac {1-0}{n}}={\frac {1}{n}}$. We also know $x_{k}=0+k\Delta x={\frac {k}{n}}.$ Let's rewrite the sum in terms of $\Delta x$ and $x_{k}$: ${\begin{aligned}\lim _{n\rightarrow \infty }\sum _{k=1}^{n}{\frac {k}{n^{2}}}{\sqrt {1-{\frac {k^{2}}{n^{2}}}}}&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}{\frac {1}{n}}{\frac {k}{n}}{\sqrt {1-{\frac {k^{2}}{n^{2}}}}}\\&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\Delta x(k\Delta x){\sqrt {1-(k\Delta x)^{2}}}\\&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}(\Delta x)x_{k}{\sqrt {1-x_{k}^{2}}}.\end{aligned}}$ In comparing what we want, $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\Delta xf(x_{k})$, with $\lim _{n\rightarrow \infty }\sum _{k=1}^{n}(\Delta x)x_{k}{\sqrt {1-x_{k}^{2}}}$, we must $f(x)=x{\sqrt {1-x^{2}}}$. Therefore $f(x)=x{\sqrt {1-x^{2}}}$.

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