MATH101 April 2011
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Question 07

Evaluate
 $\int _{1}^{2}(x1)dx$
using a limit of Riemann sums. No credit will be given for using another method (but you can use another method to check your answer). You may use the formulas
 $\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}\qquad \sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}$

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

Recall that the integral
 $\int _{a}^{b}f(x)\,dx$
is given by
 $\lim _{N\to \infty }\sum _{i=1}^{N}f\left(a+i\triangle x\right)\triangle x$
where
 $\triangle x={\frac {ba}{N}}$

Hint 2

Following from the first hint:
 $\int _{a}^{b}f(x)\,dx=\lim _{N\to \infty }\sum _{i=1}^{N}f(a+i(ba)/N){\frac {ba}{N}}$
and in this case, we have a = 1 and b = 2 for $f(x)=x1$.

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.
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Solution

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Since $f(a+i(ba)/N)=f(1+3i/N)=1+3i/N1$ we follow the hints, to obtain
 ${\begin{aligned}\int _{1}^{2}(x1)\,dx&=\lim _{N\to \infty }\sum _{i=1}^{N}f(1+3i/N){\frac {3}{N}}\\&=\lim _{N\to \infty }\sum _{i=1}^{N}(1+3i/N1){\frac {3}{N}}\\&=\lim _{N\to \infty }{\frac {3}{N}}\sum _{i=1}^{N}\left(2+{\frac {3i}{N}}\right)\\&=\lim _{N\to \infty }\left({\frac {6}{N}}\sum _{i=1}^{N}1+{\frac {9}{N^{2}}}\sum _{i=1}^{N}i\right)\end{aligned}}$
Since
 $\sum _{i=1}^{N}1=N\quad {\text{ and }}\quad \sum _{i=1}^{N}i={\frac {N(N+1)}{2}},$
we can continue our computation
 ${\begin{aligned}\int _{1}^{2}(x1)\,dx&=\lim _{N\to \infty }\left({\frac {6}{N}}N+{\frac {9}{N^{2}}}{\frac {N(N+1)}{2}}\right)\\&=6+{\frac {9}{2}}\lim _{N\to \infty }{\frac {N+1}{N}}\\&=6+{\frac {9}{2}}\\&={\frac {3}{2}}.\end{aligned}}$
You can always compare your answer to the computation of the integral directly and see that the answers match:
 ${\begin{aligned}\int _{1}^{2}(x1)\,dx&=\left.{\frac {x^{2}}{2}}x\right_{1}^{2}\\&={\frac {4}{2}}2\left({\frac {1}{2}}+1\right)\\&={\frac {3}{2}}.\end{aligned}}$

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