Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (d)

MATH105 April 2011

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Question 09 (d)

Each of the short-answer questions below is worth 5 points. Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. Determine whether the improper integral $\int _{0}^{2}{\frac {2x}{1-x^{2}}}dx$ is convergent or divergent. If convergent, write the value of the integral in the box. If not, write "divergent" and explain why in the space provided below.

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
Check to make sure that the integral makes sense. Where is the integral not defined?

Hint 2
An improper integral $\int _{a}^{b}f(x)\,dx$ with the discontinuity at the point $c$ in the interval from $a$ to $b$ converges if and only if $\lim _{t\to c^{-}}\int _{a}^{t}f(x)\,dx$ and $\lim _{s\to c^{+}}\int _{s}^{b}f(x)\,dx$ both converge.

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. We want to determine whether the integral $\int _{0}^{2}{\frac {2x}{1-x^{2}}}\,dx$ converges. We notice that the integrand is not defined at the point $x=1$ so the integral is improper. Hence, we break up the integral at this discontinuity to get that $\int _{0}^{2}{\frac {2x}{1-x^{2}}}\,dx=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {2x}{1-x^{2}}}\,dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {2x}{1-x^{2}}}\,dx$ To evaluate the first integral, we use a substitution namely $u=1-x^{2}$ and get that $du=-2xdx$. The endpoints change to $u(0)=1-(0)^{2}=1$ and $u(b)=1-b^{2}$ ${\begin{aligned}\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {2x}{1-x^{2}}}\,dx&=\lim _{b\to 1^{-}}\int _{1}^{1-b^{2}}{\frac {-du}{u}}\\&=\lim _{b\to 1^{-}}{\big (}-\ln |u|{\big )}{\big |}_{1}^{1-b^{2}}\\&=\lim _{b\to 1^{-}}-\ln(1-b^{2})+\ln(1)\end{aligned}}$ and this last limit evaluates to negative infinity. Hence, the integral diverges.

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