Answer
\begin{align}
\lim\limits_{max \varDelta x_{k} \to 0} \sum_{k=1}^{n} \sqrt {4-(x_{k}^{*})^{2}} \varDelta x_{k} = 2\pi
\end{align}
Work Step by Step
The given Riemann sum is
\begin{align}
\lim\limits_{max \varDelta x_{k} \to 0} \sum_{k=1}^{n} \sqrt {4-(x_{k}^{*})^{2}} \varDelta x_{k}
\end{align}
The definite integral form of the given Riemann sum is
\begin{alignat}{1}
\int_{-2}^{2} \sqrt {4-x^{2}} \ dx
\end{alignat}
Use trigonometric substitution x = 2 $\sin t$ $\Rrightarrow$ dx = 2 $\cos t$ dt:
\begin{alignat}{1}
&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt {4-(2\sin t)^{2}} \times 2\cos t \ dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4(\cos t)^{2} = \\ & = 2 \times \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos (2t)) \ dt = 2 \times (\pi + 0) = 2\pi
\end{alignat}
