use integration, the direct comparison test, or the limit comparison test to test the integral for…

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you prefer\n\nchoose the correct choice below.\n\noa. by the limit comparison test, diverges because and are continuous on the interval 1, ∞), and diverges.\n\nob. by the direct comparison method, converges because and are continuous on the interval 1, ∞), on 1, ∞), and converges.\n\noc. by the direct comparison method, diverges because and are continuous on the interval 1, ∞), and diverges\n\nod. the integral cannot be evaluated using integration, so the integral diverges.

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you prefer\n\nchoose the correct choice below.\n\noa. by the limit comparison test, diverges because and are continuous on the interval 1, ∞), and diverges.\n\nob. by the direct comparison method, converges because and are continuous on the interval 1, ∞), on 1, ∞), and converges.\n\noc. by the direct comparison method, diverges because and are continuous on the interval 1, ∞), and diverges\n\nod. the integral cannot be evaluated using integration, so the integral diverges.

Answer

Explanation:

Step1: Recall the Direct Comparison Test

For two functions (f(x)) and (g(x)) such that (0\leq f(x)\leq g(x)) for (x\in[a,\infty)). If (\int_{a}^{\infty}g(x)dx) converges, then (\int_{a}^{\infty}f(x)dx) converges. For (x\in[1,\infty)), we have (x^{4}+5>x^{4}), so (\frac{1}{x^{4}+5}<\frac{1}{x^{4}}).

Step2: Evaluate (\int_{1}^{\infty}\frac{1}{x^{4}}dx)

We know that (\int\frac{1}{x^{p}}dx=\frac{x^{-p + 1}}{-p+1}+C) ((p\neq1)). For (p = 4) and the improper integral (\int_{1}^{\infty}\frac{1}{x^{4}}dx=\lim_{b\rightarrow\infty}\int_{1}^{b}x^{-4}dx). [ \begin{align*} \lim_{b\rightarrow\infty}\int_{1}^{b}x^{-4}dx&=\lim_{b\rightarrow\infty}\left[\frac{x^{-4 + 1}}{-4+1}\right]{1}^{b}\ &=\lim{b\rightarrow\infty}\left[\frac{x^{-3}}{-3}\right]{1}^{b}\ &=\lim{b\rightarrow\infty}\left(-\frac{1}{3b^{3}}+\frac{1}{3}\right)\ &=\frac{1}{3} \end{align*} ] Since (\int_{1}^{\infty}\frac{1}{x^{4}}dx) converges and (0\leq\frac{1}{x^{4}+5}\leq\frac{1}{x^{4}}) for (x\in[1,\infty))

Answer:

B. By the Direct Comparison Method, (\int_{1}^{\infty}\frac{dx}{x^{4}+5}) converges because (\frac{1}{x^{4}+5}) and (\frac{1}{x^{4}}) are continuous on the interval ([1,\infty)), (0\leq\frac{1}{x^{4}+5}\leq\frac{1}{x^{4}}) on ([1,\infty)), and (\int_{1}^{\infty}\frac{1}{x^{4}}dx) converges.