Question

${\int }_0^2{x}^3\sqrt{2x-x^2}dx$ is equal to

• A. $\frac{7\pi }{2}$
• B. $\frac{7\pi }{4}$
• C. $\frac{7\pi }{8}$
• D. $\frac{7\pi }{16}$

Answer 1

Answer: (C)

$\frac{7\pi }{8}$

Let $I={\int }_0^2{x}^3\sqrt{2x-x^2}dx$

Substitute $x=2{\sin }^2\theta \Rightarrow dx=4\sin \theta \cos \theta d\theta$

$\therefore I={\int }_0^{\pi /2}8{\sin }^6\theta \sqrt{4{\sin }^2\theta -4{\sin }^4\theta }\left(4\sin \theta \cos \theta \right)d\theta$

$=64{\int }_0^{\pi /2}{\sin }^8\theta {\cos }^2\theta d\theta =64.\frac{\mathrm{7.5.3.1.1}}{\mathrm{10.8.6.4.2}}.\frac{\pi }{2}=\frac{7\pi }{8}$