MATH103 April 2014
• Q1 (a) • Q1 (b) i • Q1 (b) ii • Q1 (c) i • Q1 (c) ii • Q1 (c) iii • Q1 (d) i • Q1 (d) ii • Q1 (d) iii • Q1 (e) i • Q1 (e) ii • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (c) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q9 (c) • Q10 (a) • Q10 (b) • Q10 (c) • Q11 •
Question 10 (b)
Evaluate where .
Hint: Integration by parts!
Work must be shown for full marks. Simplify fully.
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.
Is it easier to differentiate or integrate it?
Use integration by parts and choose as the function to differentiate. What is the function that we integrate?
How can the Fundamental Theorem of Calculus help us differentiate ?
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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Following the suggestion in the question, we will integrate by parts. It is much easier to differentiate than integrate it (we'd have to integrate the integral) and so we let
The remaining part must be and so therefore . This is fairly straightforward to integrate and so we get
To differentiate we will use the fundamental theorem of calculus which states that
Applying this to our function yields
Using the information we have obtained with integration by parts formula,
The simplification comes from the fact that
for any function . In our particular case, and . The last integral can be evaluated by substitution. Let so that . If we change variables, we have to be careful with the bounds. When then and when then . Therefore the bounds don't change and the integral becomes
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