Science:Math Exam Resources/Courses/MATH103/April 2010/Question 01 (e)

MATH103 April 2010

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Question 01 (e)
Multiple Choice Question: Exactly ONE of the answers provided is correct. There is no partial credit in this questions. Which of the following improper integrals converges? i. $I=\int _{1}^{\infty }e^{3x}\,dx$ ii. $I=10^{-5}\int _{1}^{\infty }x^{-1}\,dx$ iii. $I=\int _{1}^{\infty }x^{-1/2}\,dx$ iv. $I=\int _{1}^{\infty }x^{-2}\,dx$ v. $I=\int _{1}^{\infty }{\frac {x^{2}}{1000}}\,dx$

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
For an improper integral of the form $\int _{1}^{\infty }f(x)\,dx$ to converge it is necessary (but not sufficient) that f(x) approaches 0 as x goes to infinity. Check this condition first to rule out some candidates that do not satisfy this.

Hint 2
For the remaining candidates try the integral test.

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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. Let's check first the if the integrand goes to 0, as a necessary condition for convergence. i. $\lim _{x\rightarrow \infty }e^{3x}=\infty$ ii. $\lim _{x\rightarrow \infty }x^{-1}=0\checkmark$ iii. $\lim _{x\rightarrow \infty }x^{-1/2}=0\checkmark$ iv. $\lim _{x\rightarrow \infty }x^{-2}=0\checkmark$ v. $\lim _{x\rightarrow \infty }{\frac {x^{2}}{1000}}=\infty$ This only leaves candidates ii., iii., and iv.. $\displaystyle {\text{Since }}x^{1/2}<x<x^{2}{\text{ we have }}x^{-1/2}>x^{-1}>x^{-2}{\text{ and thus }}$ $\int _{1}^{\infty }x^{-1/2}\,dx>\int _{1}^{\infty }x^{-1}\,dx>\int _{1}^{\infty }x^{-2}\,dx$ Since we are told that only one of the five integrals converges it must be iv.. (E.g., if the integral of ii. was finite, and the integral in iv. is less than the integral in ii., then iv. would also necessarily be finite.) And indeed, iv. does converge: By the integral test, $\int _{1}^{\infty }x^{-2}\,dx$ converges if $\sum _{n=1}^{\infty }n^{-2}$ converges. This sum does converge by the p-series test. Hence the correct answer is iv.. As a fun fact, $\sum _{n=1}^{\infty }n^{-2}={\frac {\pi ^{2}}{6}}$

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