175 – Quiz 5

Here is quiz 5.

Problem 1 asks for the partial fractions decomposition of {\displaystyle\frac1{x^4-x^2}.} To find this, first we factor the denominator:

{x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).} This means that there must be constants {A,B,C,D} such that

\displaystyle \frac1{x^4-x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{x+1}.

To find these constants, we add the fractions on the right hand side, and obtain

\displaystyle \frac1{x^4-x^2}=\frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.

This means that

\displaystyle 1=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1).

Setting {x=0} gives us

\displaystyle 1=-B\mbox{ or }B=-1.

Setting {x=1} gives us

\displaystyle 1=2C\mbox{ or }C=1/2.

Setting {x=-1} gives us

\displaystyle 1=-2D\mbox{ or }D=-1/2.

Setting {x=2} and using the values of {B,C,D} just found, gives us

\displaystyle 1=6A-3+6-2\mbox{ or }1=6A+1\mbox{ or }A=0.

Putting this together, the partial fractions decomposition is

\displaystyle \frac1{x^4-x^2}=-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}.

Problem 2 asks to determine whether {\displaystyle\int_1^2\frac{dx}{x^4-x^2}} converges. Note that the expression we are integrating is defined in {(1,2],} but not at {x=1,} so this is an improper integral of Type II and to evaluate it we use the definition:

\displaystyle \int_1^2\frac{dx}{x^4-x^2}=\lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}.

To evaluate this last expression, we use the result from Problem 1:

\displaystyle \begin{array}{rl} \displaystyle \lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}&=\displaystyle\lim_{a\rightarrow1^+}\int_a^2-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}\,dx\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_a^2\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac1a+\frac12\ln(a-1)-\frac12\ln(a+1)\right]. \end{array}

This expression diverges because {\displaystyle \lim_{a\rightarrow1^+}\frac1a-\frac12\ln(a+1)=1-\frac12\ln2,} but

\displaystyle \lim_{a\rightarrow1^+}\frac12\ln(a-1)=-\infty.

Problem 3 asks to determine whether {\displaystyle\int_2^\infty\frac{dx}{x^4-x^2}} converges. Note that the expression we are integrating is defined in {[2,\infty)} but, of course, the interval of integration is infinite, so this is an improper integral of Type I and to evaluate it we use the definition:

\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=\lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}.

To evaluate this last expression, we proceed as in Problem 2:

\displaystyle \begin{array}{rl} \displaystyle \lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}&=\displaystyle\lim_{N\rightarrow\infty}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_2^N\\ &=\displaystyle\lim_{N\rightarrow\infty}\left[\frac1N+\frac12\ln(N-1)-\frac12\ln(N+1)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]. \end{array}

To evaluate the expression within the first set of parentheses, we use that

\displaystyle \frac12\ln(N-1)-\frac12\ln(N+1) \displaystyle =\ln\sqrt{N-1}+\ln\left(\frac1{\sqrt{N+1}}\right) \displaystyle=\ln\left(\sqrt{\frac{N-1}{N+1}}\right).

The limit of this expression as {N\rightarrow\infty} is {\ln\sqrt1=0,} because {\displaystyle\lim_{N\rightarrow\infty}\frac{N-1}{N+1}=\lim_{N\rightarrow\infty}\frac{1}{1}=1,} using l’Hôpital’s rule. This means that

\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=-\frac12+\frac12\ln3.

(In particular, the integral converges.)

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This entry was posted on Friday, November 6th, 2009 at 3:33 pm and is filed under 175: Calculus II. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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