Question

${\int }_0^a{x}^4{(a^2-x^2)}^{1/2}dx$ equals

• A. $\frac{\pi a^5}{32}$
• B. $\frac{\pi a^6}{32}$
• C. $\frac{\pi a^2}{32}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi a^6}{32}$

$I={\int }_0^a{x}^4\sqrt{a^2-x^2}dx={\int }_0^a{x}^5\sqrt{{\left(\frac{a}{x}\right)}^2-1}dx$
Put ${\left(\frac{a}{x}\right)}^2=t\Rightarrow (-\frac{2a^2}{x^3})dx=dt$
$I=-{\int }_{\infty }^1\frac{a^8}{t^4}\sqrt{t^2-1}dt={\int }_1^{\infty }\frac{a^8\sqrt{t^2-1}}{\sqrt{t^2}}dt$
$=a^8{\int }_1^{\infty }\sqrt{1-\frac{1}{t^2}}dt=\frac{\pi a^6}{32}$