Question

Solve given definite integral.

${\int }_0^4 x^3\sqrt{x^2+4}dx$

Short answer

242.78

Step-by-step solution

Step1. Given Information

The integral is as follows.

${\int }_0^4 x^3\sqrt{x^2+4}dx$

The objective is to solve the integral.

Step2. Assumptions

To solve the integral, let $u=x^2$

Now, derivation of above equation is solved below.

$$\begin{gathered} u=x^2 \\ du=2xdx \end{gathered}$$

Step3. Solution

The integral is solved below.

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

$=\frac{1}{2}\int 2v^2\left(v^2-4\right)dv\left[v=\sqrt{u+4},dv=\frac{9}{2\sqrt{u+4}}du\right]$

$\begin{array}{r}=\int v^4-4v^{2}dv\\ =\int v^{4}dv-4\int v^{2}dv\\ =\frac{1}{5}(v{)}^5-\frac{4}{3}(v{)}^3\end{array}$

$=\frac{1}{5}{\left(x^2+4\right)}^{\frac{5}{2}}-\frac{4}{3}{\left(x^2+4\right)}^{\frac{3}{2}}$

$=\left(\frac{x^2}{5}-\frac{8}{15}\right){\left(x^2+4\right)}^{\frac{3}{2}}$

Step4. Solution

The definite integral is solved below.

${\int }_0^4 x^3\sqrt{x^2+4}dx$

$={\left[\left(\frac{x^2}{5}-\frac{8}{15}\right){\left(x^2+4\right)}^{\frac{3}{2}}\right]}_0^4$

$=\left[\left(\frac{4^2}{5}-\frac{8}{15}\right){\left(4^2+4\right)}^{\frac{3}{2}}\right]-\left[\left(\frac{0^2}{5}-\frac{8}{15}\right){\left(0^2+4\right)}^{\frac{3}{2}}\right]$

$=\frac{320\sqrt{5}}{3}+\frac{64}{15}$

$=242.78$

Therefore, the value is$242.78$