Chapter 5, Problem 1RE

Evaluate the integrals.

${\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}$

To determine

To calculat e: The value of the integral ${\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}$ .

Answer to Problem 1RE

The value of the integral ${\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}$ is $-\frac{1}{4x^2}+\frac{8}{3}{x}^{\frac{3}{2}}+C$ .

Explanation of Solution

Calcul ation:

Consider the given integral,

${\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}$

Now calculate the integral by using properties of integration as,

$\begin{array}{c}{\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}={\displaystyle \int \frac{1}{2x^3}dx+{\displaystyle \int 4\sqrt{x}}}dx\\ =\frac{1}{2}{\displaystyle \int x^{-3}dx+4{\displaystyle \int x^{\frac{1}{2}}dx}}\\ =\frac{1}{2}\left[\frac{x^{-3+1}}{-3+1}\right]+4\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+C\\ =\frac{1}{2}\left[\frac{x^{-2}}{-2}\right]+4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]+C\end{array}$

Further solve as,

$\begin{array}{c}{\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}=-\frac{1}{4x^2}+4\cdot \frac{2}{3}{x}^{\frac{3}{2}}+C\\ =-\frac{1}{4x^2}+\frac{8}{3}{x}^{\frac{3}{2}}+C\end{array}$

Therefore, the value of the integral ${\displaystyle \int \left[\frac{1}{2x^3}+4\sqrt{x}\right]dx}$ is $-\frac{1}{4x^2}+\frac{8}{3}{x}^{\frac{3}{2}}+C$ .