Science:Math Exam Resources/Courses/MATH103/April 2009/Question 01 (a)

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MATH103 April 2009

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  

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Question 01 (a)

Multiple choice question: Select ONE correct answer. You will not be graded for any work. Which of these integrals does the sum $\sum _{k=1}^{N}\left({\frac {3}{N}}\right)\left[\left(-2+{\frac {3k}{N}}\right)^{3}-1\right]$ approximate as $N\rightarrow \infty$? (a) $\int _{0}^{3}x^{3}\,dx$ (b) $\int _{-2}^{1}(x^{3}-1)\,dx$ (c) $\int _{0}^{3}(x^{3}-2)\,dx$ (d) $\int _{-2}^{1}x^{3}\,dx$ (e) $\int _{0}^{1}(x-2)^{3}\,dx$

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
The formula for the limit of the Rieman sum is $\lim _{N\rightarrow \infty }\sum _{k=1}^{N}\left({\frac {b-a}{N}}\right)\cdot f\left(a+{\frac {k(b-a)}{N}}\right)=\int _{a}^{b}f(x)dx.$

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Solution

As we see in the formula for the limit of the Riemann sum $\lim _{N\rightarrow \infty }\sum _{k=1}^{N}\left({\frac {b-a}{N}}\right)\cdot f\left(a+{\frac {k(b-a)}{N}}\right)=\int _{a}^{b}f(x)dx$ the factor that is devided by $\ N$, must be $\ b-a$. So, we know that $\ b-a=3$. Now we search for the variable $\displaystyle x_{k}=a+{\frac {k(b-a)}{N}}=a+{\frac {3k}{N}}.$ The only term, that matches this expression is $\ -2+{\frac {3k}{N}}$ and thus a must be equal to -2. With 3=b-a = b+2 we find that b=1. The last thing is to find the function f. From the formula of the limit of the Riemann sum we know that $f\left(a+{\frac {k(b-a)}{N}}\right)=\left[\left(a+{\frac {k(b-a)}{N}}\right)^{3}-1\right],$ $f\left(-2+{\frac {3k}{N}}\right)=\left(-2+{\frac {3k}{N}}\right)^{3}-1,$ $\ f(x)=x^{3}-1.$ So the answer is (b), $\lim _{N\rightarrow \infty }\sum _{k=1}^{N}\left({\frac {3}{N}}\right)\cdot \left[\left(-2{\frac {k3}{N}}\right)^{3}-1\right]=\int _{-2}^{1}(x^{3}-1)dx.$

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