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

MATH105 April 2011

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[hide] 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.

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[show] Hint 1
Check to make sure that the integral makes sense. Where is the integral not defined?

[show] 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.

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[show] Solution
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.