Science:Math Exam Resources/Courses/MATH101/April 2011/Question 07

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MATH101 April 2011

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Question 07
Evaluate $\int _{-1}^{2}(x-1)dx$ 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 $\triangle x={\frac {b-a}{N}}$

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

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

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

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