find the area bounded by y = (x - 11)/(x^2 - 2x - 63), x = 2, x = 5, and y = 0. (round the answer to four…

Explanation:

Step1: Factor the denominator

First, factor $x^{2}-2x - 63=(x - 9)(x+7)$. Then we use partial - fraction decomposition for $\frac{x - 11}{(x - 9)(x + 7)}$. Let $\frac{x - 11}{(x - 9)(x + 7)}=\frac{A}{x - 9}+\frac{B}{x + 7}$. Cross - multiply gives $x-11=A(x + 7)+B(x - 9)$. Set $x = 9$, then $9-11=A(9 + 7)+B(0)$, so $A=-\frac{1}{8}$. Set $x=-7$, then $-7-11=A(0)+B(-7 - 9)$, so $B=\frac{9}{8}$. So $\frac{x - 11}{x^{2}-2x - 63}=-\frac{1}{8(x - 9)}+\frac{9}{8(x + 7)}$.

Step2: Calculate the definite integral

The area $A=\int_{2}^{5}\left| \frac{x - 11}{x^{2}-2x - 63}\right|dx=\int_{2}^{5}\left(-\frac{1}{8(x - 9)}+\frac{9}{8(x + 7)}\right)dx$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have: [ \begin{align*} \int_{2}^{5}\left(-\frac{1}{8(x - 9)}+\frac{9}{8(x + 7)}\right)dx&=-\frac{1}{8}\int_{2}^{5}\frac{1}{x - 9}dx+\frac{9}{8}\int_{2}^{5}\frac{1}{x + 7}dx\ &=-\frac{1}{8}[\ln|x - 9|]{2}^{5}+\frac{9}{8}[\ln|x + 7|]{2}^{5}\ &=-\frac{1}{8}(\ln|5 - 9|-\ln|2 - 9|)+\frac{9}{8}(\ln|5 + 7|-\ln|2+7|)\ &=-\frac{1}{8}(\ln4-\ln7)+\frac{9}{8}(\ln12-\ln9)\ &=-\frac{1}{8}\ln\frac{4}{7}+\frac{9}{8}\ln\frac{12}{9}\ &=-\frac{1}{8}\ln\frac{4}{7}+\frac{9}{8}\ln\frac{4}{3} \end{align*} ]

Step3: Simplify and round the result

[ \begin{align*} -\frac{1}{8}\ln\frac{4}{7}+\frac{9}{8}\ln\frac{4}{3}&=\frac{1}{8}\left(-\ln\frac{4}{7}+9\ln\frac{4}{3}\right)\ &=\frac{1}{8}\left(-(\ln4-\ln7)+9(\ln4-\ln3)\right)\ &=\frac{1}{8}\left(-\ln4+\ln7 + 9\ln4-9\ln3\right)\ &=\frac{1}{8}\left(8\ln4+\ln7-9\ln3\right)\ &\approx\frac{1}{8}(8\times1.3863 + 1.9459-9\times1.0986)\ &=\frac{1}{8}(11.0904+1.9459 - 9.8874)\ &=\frac{1}{8}(3.1489)\ &=0.3936 \end{align*} ]

Answer:

$0.3936$