Answer
\begin{align}
\int_{0}^{10}\sqrt {10x-x^{2}} \ dx = 12.5\pi
\end{align}
Work Step by Step
The given integral is
\begin{align}
\int_{0}^{10}\sqrt {10x-x^{2}} \ dx
\end{align}
Let us solve it by completing the square and applying substitutions/geometric formulas:
\begin{alignat}{2}
&\int_{0}^{10}\sqrt {10x-x^{2}}dx=\int_{0}^{10}\sqrt {-(x^{2}-10x+25)+25} \ dx=\\ &=\int_{0}^{10}\sqrt {-(x^{2}-10x+25)+25} \ dx = \int_{0}^{10}\sqrt {-(x-5)^{2}+25} \ dx
\end{alignat}
Let u = x - 5 $\Rrightarrow$ du = dx
\begin{alignat}{1}
\int_{-5}^{5}\sqrt {-u^{2}+25} \ du
\end{alignat}
Now, use trigonometric substitution u = 5 $\sin t$ $\Rrightarrow$ du = 5 $\cos t$ dt. Thus,
\begin{alignat}{2}
& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt {25-25(\sin t)^2} \times 5\cos t \ dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 25 (\cos t)^{2} \ dt = \\ &=12.5 \times \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos (2t)) \ dt = 12.5 \times (\pi + 0) = 12.5\pi
\end{alignat}
