Chapter 5.8, Problem 16E

To determine

To evaluate the integral ${\displaystyle {\int }_0^2{x}^3}\sqrt{4x^2-x^4}dx$

Answer to Problem 16E

The value of the integral ${\displaystyle {\int }_0^2{x}^3}\sqrt{4x^2-x^4}dx$is $2\pi \simeq 6.28318$

Explanation of Solution

Given information:

The given expression is ${\displaystyle {\int }_0^2{x}^3}\sqrt{4x^2-x^4}dx$

Formula used:

Integration by parts, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$

Integrals involving $2au-u^2,a>0$

  ${\displaystyle \int u\sqrt{2au-u^2}du=\frac{2u^2-au-3a^2}{6}\sqrt{2au-u^2}+\frac{a^3}{2}{\cos }^{-1}\left(\frac{a-u}{a}\right)+c}$

Calculation: Consider, ${\displaystyle {\int }_0^2{x}^3}\sqrt{4x^2-x^4}dx$

Apply u substitution $u=x^2\Rightarrow du=2xdx\Rightarrow \frac{1}{2}du=xdx$ ; $\frac{1}{3}{u}^{-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}du=dx$

When $\begin{array}{l}x=0\Rightarrow u=0\\ x=2\Rightarrow u=4\end{array}$

The integral is simplified to ${\displaystyle {\int }_0^2{x}^3}\sqrt{4x^2-x^4}dx={\displaystyle \underset{0}{\overset{2}{\int }}{x}^2\sqrt{\left(4x^2-x^4\right)}xdx}$

Substitute for u,

  $={\displaystyle {\int }_0^4\frac{u\sqrt{4u-u^2}}{2}}du$

Take the constant out: ${\displaystyle \int a}\cdot f\left(x\right)dx=a\cdot {\displaystyle \int f}\left(x\right)dx$

  $=\frac{1}{2}{\displaystyle \underset{0}{\overset{4}{\int }}u\sqrt{2\cdot 2u-u^2}}du$

Now using the formula,

  ${\displaystyle \int u\sqrt{2au-u^2}du=\frac{2u^2-au-3a^2}{6}\sqrt{2au-u^2}+\frac{a^3}{2}{\cos }^{-1}\left(\frac{a-u}{a}\right)+c}$

  $\begin{array}{l}\frac{1}{2}{{\displaystyle \underset{0}{\overset{4}{\int }}u\sqrt{2\cdot 2u-u^2}du=\frac{1}{2}\left[\frac{2u^2-2u-3\times 2^2}{6}\sqrt{4u-u^2}+\frac{2^3}{2}{\mathrm{Cos}}^{-1}\left(\frac{2-u}{2}\right)\right]}}_0^4\\ \text{ =}\frac{1}{2}{\left[\frac{u^2-u-6}{3}\sqrt{4u-u^2}+4{\mathrm{Cos}}^{-1}\left(1-\frac{u}{2}\right)\right]}_0^4\\ \text{ =}\frac{1}{2}\left[\frac{16-4-6}{3}\sqrt{16-16}+4{\mathrm{Cos}}^{-1}\left(1+2\right)-\left(\frac{0-0-6}{3}\sqrt{0-0}+4{\mathrm{Cos}}^{-1}\left(1-0\right)\right)\right]\\ \text{ =}\frac{1}{4}\left[4{\mathrm{Cos}}^{-1}\left(-1\right)\right]\\ \text{ =}\frac{1}{2}\left[4\times \pi \right]\\ \text{ =2}\pi \simeq \text{6}\text{.28318}\end{array}$