Science:Math Exam Resources/Courses/MATH103/April 2013/Question 02 (c)

MATH103 April 2013

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Question 02 (c)
Integration: Short Answer Problems - Evaluate the following integrals: state if a definite integral does not exist; use limits for improper integrals. (full marks for correct answer; work must be shown for partial marks) $\displaystyle I_{c}=\int _{0}^{2}{\frac {1}{x^{2}-1}}\,dx$

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
Watch out! It's a trap! This integral is actually improper! First start out by splitting up at the singularity in the interval.

Hint 2
Try integration by partial fractions.

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 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. First, note that the integral above is improper with a singularity at $\displaystyle x=1$. Splitting up the integral there, we see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\\end{aligned}}$ Now, let $\displaystyle {\frac {1}{x^{2}-1}}={\frac {A}{x-1}}+{\frac {B}{x+1}}$ by the partial fraction decomposition. This gives via cross multiplication $\displaystyle 1=A(x+1)+B(x-1)$ METHOD 1: Plugging in $\displaystyle x=-1$ shows that $\displaystyle B=-{\frac {1}{2}}$ and plugging in $\displaystyle x=1$ shows that $\displaystyle A={\frac {1}{2}}$. METHOD 2: Expanding the right yields $\displaystyle 1=A-B+(A+B)x$ Since the coefficients of the constant term are equal and the coefficients of the x term must be equal, we have $\displaystyle A-B=1$ and $\displaystyle A+B=0$ Adding the two equations gives $\displaystyle 2A=1$ so $\displaystyle A={\frac {1}{2}}$ as above and subtracting the two equations gives $\displaystyle -2B=1$ and so $\displaystyle B={\frac {-1}{2}}$. Now that we have the coefficients, we note that $\displaystyle {\frac {1}{x^{2}-1}}={\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}$ and go back to the integral to see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}\left({\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}\right)\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}\left({\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}\right)\ dx\\&=\lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln |x-1|-{\tfrac {1}{2}}\ln |x+1|{\big |}_{0}^{b}+\lim _{a\to 1^{+}}{\tfrac {1}{2}}\ln |x-1|-{\tfrac {1}{2}}\ln |x+1|{\big |}_{a}^{2}\\&=\lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln |b-1|-{\tfrac {1}{2}}\ln |b+1|-({\tfrac {1}{2}}\ln |1|-{\tfrac {1}{2}}\ln |1|)\\&\quad +\lim _{a\to 1^{+}}{\tfrac {1}{2}}\ln |1|-{\tfrac {1}{2}}\ln |3|-({\tfrac {1}{2}}\ln |a-1|-{\tfrac {1}{2}}\ln |a+1|)\\\end{aligned}}$ Since $\displaystyle \lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln |b-1|$ diverges, we have that the entire integral diverges and thus the integral does not exist.

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. First, note that the integral above is improper with a singularity at $\displaystyle x=1$. Splitting up the integral there, we see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\\end{aligned}}$ Now, let $\displaystyle {\frac {1}{x^{2}-1}}={\frac {A}{x-1}}+{\frac {B}{x+1}}$ by the partial fraction decomposition. We note that ${\begin{aligned}\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {A}{x-1}}\ dx+\int _{0}^{1}{\frac {B}{x+1}}\ dx\end{aligned}}$ and ${\begin{aligned}\int _{0}^{1}{\frac {A}{x-1}}\ dx\end{aligned}}$ doesn't seem to converge since it behaves like $\int _{-1}^{0}{\frac {1}{x}}\ dx$, so the whole integral does not converge. (In particular, we don't even need to worry about what $\displaystyle A$ and $\displaystyle B$ are, since they are non-zero constants!) Let's prove this rigorously. ${\begin{aligned}\int _{0}^{1}{\frac {A}{x-1}}\ dx&=A\cdot \lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x-1}}\ dx\\&=A\cdot \lim _{b\to 1^{-}}\ln |x-1|_{0}^{b}\\&=A\cdot \left(\lim _{b\to 1^{-}}\ln |b-1|-\ln |0-1|\right)\\\end{aligned}}$ Since $\displaystyle \lim _{b\to 1^{-}}\ln |b-1|$ diverges, we have that the integral diverges and thus the entire integral does not exist.

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