evaluate the following integral or state that it diverges.\n int_{2}^{infty}\frac{dx}{x^{4}} \nselect the…

evaluate the following integral or state that it diverges.\n int_{2}^{infty}\frac{dx}{x^{4}} \nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. the integral converges and (int_{2}^{infty}\frac{dx}{x^{4}}=) . (type an exact answer.)\nb. the integral diverges.

evaluate the following integral or state that it diverges.\n int_{2}^{infty}\frac{dx}{x^{4}} \nselect the correct choice and, if necessary, fill in the answer box to complete your choice.\na. the integral converges and (int_{2}^{infty}\frac{dx}{x^{4}}=) . (type an exact answer.)\nb. the integral diverges.

Answer

Explanation:

Step1: Recall power - rule for integration

The antiderivative of $x^n$ is $\frac{x^{n + 1}}{n+1}+C$ for $n\neq - 1$. For the function $f(x)=\frac{1}{x^{4}}=x^{-4}$, its antiderivative $F(x)=\frac{x^{-4 + 1}}{-4+1}=-\frac{1}{3x^{3}}$.

Step2: Evaluate the improper integral

The improper integral $\int_{2}^{\infty}\frac{dx}{x^{4}}=\lim_{b\rightarrow\infty}\int_{2}^{b}x^{-4}dx$. First, find $\int_{2}^{b}x^{-4}dx=\left[-\frac{1}{3x^{3}}\right]{2}^{b}=-\frac{1}{3b^{3}}+\frac{1}{3\times2^{3}}$. Then, find the limit $\lim{b\rightarrow\infty}\left(-\frac{1}{3b^{3}}+\frac{1}{24}\right)$. As $b\rightarrow\infty$, $\lim_{b\rightarrow\infty}\frac{-1}{3b^{3}} = 0$. So, $\lim_{b\rightarrow\infty}\left(-\frac{1}{3b^{3}}+\frac{1}{24}\right)=\frac{1}{24}$.

Answer:

A. The integral converges and $\int_{2}^{\infty}\frac{dx}{x^{4}}=\frac{1}{24}$