the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{-6}…

the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{-6} \frac{50}{x^{2}-25} d x \n int_{-infty}^{-6} \frac{50}{x^{2}-25} d x=quad \text { (type an exact answer.) }

the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{-6} \frac{50}{x^{2}-25} d x \n int_{-infty}^{-6} \frac{50}{x^{2}-25} d x=quad \text { (type an exact answer.) }

Answer

Explanation:

Step1: Decompose the integrand

Use partial fraction decomposition on (\frac{50}{x^{2}-25}=\frac{50}{(x - 5)(x + 5)}). Let (\frac{50}{(x - 5)(x + 5)}=\frac{A}{x - 5}+\frac{B}{x + 5}). Then (50=A(x + 5)+B(x - 5)). Set (x = 5), we get (50=A(5 + 5)+B(5 - 5)), so (A = 5). Set (x=-5), we get (50=A(-5 + 5)+B(-5 - 5)), so (B=-5). Thus (\frac{50}{x^{2}-25}=\frac{5}{x - 5}-\frac{5}{x + 5}).

Step2: Calculate the improper integral

(\int_{-\infty}^{-6}\frac{50}{x^{2}-25}dx=\lim_{a\rightarrow-\infty}\int_{a}^{-6}(\frac{5}{x - 5}-\frac{5}{x + 5})dx) (=\lim_{a\rightarrow-\infty}[5\ln|x - 5|-5\ln|x + 5|]{a}^{-6}) (=\lim{a\rightarrow-\infty}[5\ln|\frac{x - 5}{x + 5}|]{a}^{-6}) Substitute the upper and lower limits: (=5\ln|\frac{-6 - 5}{-6 + 5}|-5\lim{a\rightarrow-\infty}\ln|\frac{a - 5}{a + 5}|) Since (\lim_{a\rightarrow-\infty}\frac{a - 5}{a + 5}=\lim_{a\rightarrow-\infty}\frac{1-\frac{5}{a}}{1+\frac{5}{a}} = 1), and (\ln1 = 0) (=5\ln11-5\times0)

Answer:

(5\ln11)