MATH103 April 2016
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) (i) • Q6 (a) (ii) • Q6 (a) (iii) • Q6 (a) (iv) • Q6 (a) (v) • Q6 (b) (i) • Q6 (b) (ii) • Q6 (b) (iii) • Q7 (a) (i) • Q7 (a) (ii) • Q7 (b) (i) • Q7 (b) (ii) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) •
Question 01 (a)
Express as a definite integral of the form . Do not evaluate the integral.
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!
Recall that , where and .
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 .
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We write the given series as a right Riemann sum, i.e., , and try to find and , and in turn .
The given integral has lower and upper bounds , respectively, so , which accounts for the first factor of each term in the series.
Therefore, the second factor must be .
We know that , so we can write in terms of as follows:
which implies that , whence . Therefore