Science:Math Exam Resources/Courses/MATH101/April 2012/Question 01 (j)
• 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) • Q3 (c) • Q3 (d) • Q4 (a) • Q4 (c) • Q5 • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) •
Question 01 (j) |
---|
Short-Answer Question. Show all your work, simplify your answer as much as possible. Using a Maclaurin series, the number is found to be an approximation for . Give the best upper bound you can for . |
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 |
---|
Direct integration is not an option because the anti-derivative of x4e-x2 can not be calculated. Instead, start by writing the integrand as a series and then integrate term-by-term. Recall that |
Hint 2 |
---|
The nature of the resulting series representation of the integral then suggests a certain theorem to find your error bound. |
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 |
---|
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. The Maclauren series of the integrand is found by: Therefore, integrating term-by-term gives: This confirms the given approximation, Moreover, this shows that the integral can be represented by an alternating series in the form Since the positive numbers an are evidently monotonically decreasing, the Alternating Series Estimation Theorem then says that an error bound of a partial sum can be given by the next term ak+1. In this case, the statement translates to the bound: |