Question

$\underset{0}{\overset{2a}{\int }}{x}^3\sqrt{2ax-x^2}dx$ is equal to

• A. $\frac{7}{4}{a}^5$
• B. $\frac{7\pi }{8}{a}^5$
• C. $\frac{7\pi }{8}{a}^4$
• D. $\frac{7\pi }{8}{a}^3$

Answer 1

The correct option is B $\frac{7\pi }{8}{a}^5$

Given, $I=\underset{0}{\overset{2a}{\int }}{x}^3\sqrt{2ax-x^2}dx$
$\Rightarrow I=\underset{0}{\overset{2a}{\int }}{x}^3\cdot x^{\frac{1}{2}}\sqrt{2a-x}dx$
Let, $x=2a{\sin }^2\theta$
$\Rightarrow dx=4a\sin \theta \cos \theta d\theta \Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(2a{\sin }^2\theta {)}^{\frac{7}{2}}\cdot \sqrt{2a}\cos \theta \cdot 4a\sin \theta \cos \theta d\theta \Rightarrow I=64a^5\underset{0}{\overset{\pi /2}{\int }}{\sin }^8\theta \cdot {\cos }^2\theta d\theta \Rightarrow I_{8,2}=64a^5\cdot \frac{(7\cdot 5\cdot 3\cdot 1)\left(1\right)}{10\cdot 8\cdot 6\cdot 4\cdot 2}\cdot \frac{\pi }{2}\{\because I_{m,n}=\frac{(m-1)(m-3)(m-5)\cdots (n-1)(n-3)(n-5)}{(m+n)(m+n-2)(m+n-4)\cdots }\cdot \frac{\pi }{2}\text{, when both}m,n\text{is even}\}I=\frac{7\pi }{8}{a}^5$