the integral in this exercise converges. evaluate the integral without using a table.\n int_{11}^{infty}…

the integral in this exercise converges. evaluate the integral without using a table.\n int_{11}^{infty} \frac{2}{v^{2}-v} d v \n int_{11}^{infty} \frac{2}{v^{2}-v} d v=quad \text { (use parentheses to clearly denote the argument of each function.) }

the integral in this exercise converges. evaluate the integral without using a table.\n int_{11}^{infty} \frac{2}{v^{2}-v} d v \n int_{11}^{infty} \frac{2}{v^{2}-v} d v=quad \text { (use parentheses to clearly denote the argument of each function.) }

Answer

Explanation:

Step1: Decompose the integrand

We use partial fraction decomposition. Given (\frac{2}{v^{2}-v}=\frac{2}{v(v - 1)}). Let (\frac{2}{v(v - 1)}=\frac{A}{v}+\frac{B}{v - 1}). Then (2=A(v - 1)+Bv). Set (v = 0), we get (A=-2). Set (v = 1), we get (B = 2). So (\frac{2}{v^{2}-v}=\frac{-2}{v}+\frac{2}{v - 1}).

Step2: Evaluate the improper integral

(\int_{11}^{\infty}\frac{2}{v^{2}-v}dv=\lim_{b\rightarrow\infty}\int_{11}^{b}(\frac{-2}{v}+\frac{2}{v - 1})dv) [ \begin{align*} \lim_{b\rightarrow\infty}\int_{11}^{b}(\frac{-2}{v}+\frac{2}{v - 1})dv&=\lim_{b\rightarrow\infty}\left[-2\ln|v|+2\ln|v - 1|\right]{11}^{b}\ &=\lim{b\rightarrow\infty}\left[2\ln\left(\frac{v - 1}{v}\right)\right]{11}^{b}\ &=\lim{b\rightarrow\infty}\left(2\ln\left(\frac{b - 1}{b}\right)-2\ln\left(\frac{11 - 1}{11}\right)\right) \end{align*} ] Since (\lim_{b\rightarrow\infty}\frac{b - 1}{b}=\lim_{b\rightarrow\infty}(1-\frac{1}{b}) = 1) and (\ln(1)=0) [ \begin{align*} &=0-2\ln\left(\frac{10}{11}\right)\ &=2\ln\left(\frac{11}{10}\right) \end{align*} ]

Answer:

(2\ln\left(\frac{11}{10}\right))