# Question #07106

Nov 7, 2017

$\frac{3}{\sqrt[3]{5}}$

#### Explanation:

The first step is to solve the integral:

$\int \frac{1}{x} ^ \left(\frac{4}{3}\right) \mathrm{dx}$ really is $\int$ ${x}^{-} \left(\frac{4}{3}\right) \mathrm{dx}$

$\int$ ${x}^{-} \left(\frac{4}{3}\right) \mathrm{dx} \to - \frac{3}{\sqrt[3]{x}} + \text{c}$

Knowing this we can proceed:

${\int}_{5}^{\infty} \frac{1}{x} ^ \left(\frac{4}{3}\right) \mathrm{dx}$ is equal to what is below

${\lim}_{b \to \infty} - \frac{3}{\sqrt[3]{x}}$ Evaluated from $5$ to $b$

${\lim}_{b \to \infty} \left[- \frac{3}{\sqrt[3]{b}} - \left(- \frac{3}{\sqrt[3]{5}}\right)\right]$

${\lim}_{b \to \infty} \left[- \frac{3}{\sqrt[3]{b}} + \frac{3}{\sqrt[3]{5}}\right]$

When we take the limit of:

${\lim}_{b \to \infty} \left[- \frac{3}{\sqrt[3]{b}}\right]$ We get $- \frac{3}{\infty}$ Anything over infinite is zero.

$\left[- \frac{3}{\sqrt[3]{\infty}} + \frac{3}{\sqrt[3]{5}}\right]$

The answer should look like this once you take the limit of course you can disregard the zero:

$\left[0 + \frac{3}{\sqrt[3]{5}}\right]$