the integral in this exercise converges. evaluate the integral without using a table.\n int_{8}^{infty}…

Explanation:

Step1: Recall the integral formula

The integral formula for (\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{- 1}(\frac{x}{a})+C). Here (a = 8) (since (x^{2}+64=x^{2}+8^{2})), and we are dealing with an improper integral (\int_{8}^{\infty}\frac{dx}{x^{2}+64}=\lim_{b\rightarrow\infty}\int_{8}^{b}\frac{dx}{x^{2}+64}).

Step2: Evaluate the definite integral

Using the formula (\int_{8}^{b}\frac{dx}{x^{2}+64}=\left[\frac{1}{8}\tan^{-1}(\frac{x}{8})\right]_{8}^{b}). By the fundamental theorem of calculus (F(b)-F(8)), where (F(x)=\frac{1}{8}\tan^{-1}(\frac{x}{8})). So (F(b)-F(8)=\frac{1}{8}\tan^{-1}(\frac{b}{8})-\frac{1}{8}\tan^{-1}(1)).

Step3: Take the limit as (b\rightarrow\infty)

We know that (\lim_{b\rightarrow\infty}\tan^{-1}(\frac{b}{8})=\frac{\pi}{2}) and (\tan^{-1}(1)=\frac{\pi}{4}). Substitute these values into (\lim_{b\rightarrow\infty}\left(\frac{1}{8}\tan^{-1}(\frac{b}{8})-\frac{1}{8}\tan^{-1}(1)\right)). [ \begin{align*} \lim_{b\rightarrow\infty}\left(\frac{1}{8}\tan^{-1}(\frac{b}{8})-\frac{1}{8}\tan^{-1}(1)\right)&=\frac{1}{8}\lim_{b\rightarrow\infty}\tan^{-1}(\frac{b}{8})-\frac{1}{8}\tan^{-1}(1)\ &=\frac{1}{8}\times\frac{\pi}{2}-\frac{1}{8}\times\frac{\pi}{4}\ &=\frac{\pi}{16}-\frac{\pi}{32} \end{align*} ]

Step4: Simplify the expression

(\frac{\pi}{16}-\frac{\pi}{32}=\frac{2\pi - \pi}{32}=\frac{\pi}{32})

Answer:

(\frac{\pi}{32})