Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (a)

MATH103 April 2012

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Question 02 (a)
Short answer question. Simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem). Use the integral comparison test to show that $\int _{1}^{\infty }{\frac {e^{-x}}{2x-1}}\,dx$ converges.

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
Use the integral comparison test, i.e. find an integrand g(x) such that Failed to parse (syntax error): {\displaystyle \frac{e^{-x}}{2x-1} \leq g(x) \quad \text{and} \quad \int_1^\infty g(x)\,dx } converges.

Hint 2
What can you say about 1/(2x-1) on the interval on which the integration is considered?

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. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. On the interval $[1,\infty )$ we have that 2x-1 ≥ 1 and so ${\frac {e^{-x}}{2x-1}}\leq e^{-x}$ for all x ≥ 1. Therefore $\int _{1}^{\infty }{\frac {e^{-x}}{2x-1}}\,dx\leq \int _{1}^{\infty }e^{-x}\,dx$ by the integral comparison test. Since the latter integral $\int _{1}^{\infty }e^{-x}\,dx$ converges (see lecture notes or direct calculation), the former integral also converges, by the integral comparison test.

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