Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 46

Find the definite integral $\displaystyle \int^a_0 x \sqrt{a^2 - x^2} dx$

Let $u = a^2 - x^2$, then $du = -2x dx$, so $\displaystyle xdx = \frac{-du}{2}$. When $x = 0, u =a^2$ and when $x = a, u = 0$. Thus,



$
\begin{equation}
\begin{aligned}

\int^a_0 x \sqrt{a^2 - x^2} dx =& \int^a_0 \sqrt{a^2 - x^2} xdx
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \int^a_0 \sqrt{u} - \frac{du}{2}
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \frac{-1}{2} \int^a_0 u^{\frac{1}{2}} du
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \frac{-1}{2} \left[ \frac{u^{\frac{1}{2} + 1}}{\displaystyle \frac{1}{2} + 1} \right]^a_0
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \frac{-1}{2} \left[ \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right]^a_0
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \left[ \frac{-1}{\cancel{2}} \cdot \frac{\cancel{2}u^{\frac{3}{2}}}{3} \right]^a_0
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \left[ \frac{-u^{\frac{3}{2}}}{3} \right]^a_0
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \frac{-a^{\frac{3}{2}}}{3} - \frac{(-0)^{\frac{3}{2}}}{3}
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\int^a_0 x \sqrt{a^2 - x^2} dx =& \frac{-a^{\frac{3}{2}}}{3}

\end{aligned}
\end{equation}
$