Science:Math Exam Resources/Courses/MATH101/April 2011/Question 01 (a)
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q7 • Q8 •
Question 01 (a) 

For this question, the exam asks to put your answer in the box provided but also to show your work. This question is worth 3 marks. Full marks is given for correct answers placed in the box and at most 1 mark is given for an incorrect answer. You are required to simplify your answers as much as possible. Evaluate the definite 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 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 

The derivative of is almost . 
Hint 2 

Use a change of variable. 
Hint 3 

Use the change of variable . 
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.

Solution 1 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. We notice that the term is almost the derivative of what is inside of the square root (up to a constant term) and so we will use a change of variable. Let . Then we have: and hence or to match what we have in our integral: Since this is a definite integral, we also need to change the bounds of integration. If then and if then . This allows us to write: Which is an easy integral to compute: 
Solution 2 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. For such an integral, you do not have to do the change of variable if you can directly figure out the correct antiderivative. Since you have the term outside of the square root, you expect the antiderivative to be something like: Taking the derivative of this, you will see how to modify the constants to make it work: And so we know that an antiderivative of will be . Which allows us to conclude: 