Question

Solve given definite integrals.

${\int }_0^{4}x\sqrt{x^2+4}dx$

Short answer

$\frac{40\sqrt{5}}{3}-\frac{8}{3}\text{.}$

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_0^4 x\sqrt{x^2+4}dx$

The objective is to solve the integral.

Step2. Assumption

The integral is solved below.

$\int x\sqrt{x^2+4}dx$

$=\int \frac{\sqrt{u}}{2}du\left[u=x^2+4,du=2xdx\right]$

$\begin{array}{r}=\frac{1}{2}\int \sqrt{u}du\\ =\frac{u^{\frac{3}{2}}}{3}\\ =\frac{{\left(x^2+4\right)}^{\frac{3}{2}}}{3}\\ =\frac{1}{3}{\left(x^2+4\right)}^{\frac{3}{2}}\end{array}$

Step3. Solution

The definite integral is solved below.

$\begin{array}{r}{\int }_0^4 x\sqrt{x^2+4}dx\\ ={\left[\frac{1}{3}{\left(x^2+4\right)}^{\frac{3}{2}}\right]}_0^4\\ =\left[\frac{1}{3}{\left(4^2+4\right)}^{\frac{3}{2}}\right]-\left[\frac{1}{3}{\left(0^2+4\right)}^{\frac{3}{2}}\right]\\ =\frac{40\sqrt{5}}{3}-\frac{8}{3}\end{array}$

Therefore, the value is$\frac{40\sqrt{5}}{3}-\frac{8}{3}\text{.}$