Answer
$\int_{0}^{a}x~\sqrt{a^2-x^2}~dx = \frac{a^3}{3}$
Work Step by Step
$\int_{0}^{a}x~\sqrt{a^2-x^2}~dx$
Let $u = a^2-x^2$
$\frac{du}{dx} = -2x$
$dx = -\frac{du}{2x}$
When $x = 0$, then $u =a^2$
When $x = a$, then $u = 0$
$\int_{a^2}^{0} (x~\sqrt{u})\cdot (-\frac{du}{2x})$
$=\int_{0}^{a^2} \frac{1}{2} \sqrt{u}~du$
$= \frac{1}{2}~(\frac{2}{3}u^{3/2}~\vert_{0}^{a^2})$
$= \frac{1}{3}~(u^{3/2}~\vert_{0}^{a^2})$
$= \frac{1}{3}~[~(a^2)^{3/2}-(0)^{3/2}~]$
$= \frac{1}{3}~(a^3-0)$
$= \frac{a^3}{3}$
