Calculate $\int_{0}^{3}\sqrt{4-x^2}\,dx$ using a Riemann sum What is the Riemann intergral of $\sqrt{4-x^2}$ from $0$ to $3$?
I tried to write down the Riemann sum $\sum_{i=1}^n\sqrt{4-\frac{(3i)^2}{n^2}}\frac{3}{n}$. Then I tried to take limit as $n \to \infty$, but I don't know how to evaluate it. I don't know what to do with the square root. 

Building on (Unfortunately, it does not fit in the comment section) from Integrand and Dark Malthrop). Hence posting as a full version.
We can start with the series expansion,
\begin{eqnarray*}
\sqrt{a^{2}-x^{2}} &=& \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m}  x^{2 m}} 
\end{eqnarray*}
\begin{eqnarray*}
\int_{}^{}{\sqrt{a^{2}-x^{2}} dx}&=&\int { \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m}  x^{2 m}}  dx} \\
&=&  \sum_{m=0}^{\infty}{\int {(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m}  x^{2 m}}  dx} \\
&=&  \sum_{m=0}^{\infty}{(-1)^{m} a^{1-2m} \binom{\frac{1}{2}}{m}  \frac{x^{2 m+1}}{2m+1} }  \\
&=& \frac{1}{2} \left( a x \sqrt{1-\frac{x^{2}}{a^{2}} }  + a^{2} \sin^{-1}\left(\frac{x}{a}\right) \right)
\end{eqnarray*}
With $a=2$, this get to,
\begin{eqnarray*}
\int{\sqrt{4-x^{2}} dx} &=& \left(  x \sqrt{1-\frac{x^{2}}{4 } }  + 2 \sin^{-1} \frac{x}{2} \right)
\end{eqnarray*}
I don't see a reason why we cannot extend the limit beyond 2, aside from the fact that, it goes outside $\mathbb{R}$.  As mentioned by many people here, when the upper limit is $2$, it becomes to $ \left(  2 \sqrt{1-\frac{2^{2}}{4 } }  + 2 \sin^{-1} \frac{2}{2} \right) =\pi$. If the limit is indeed as posited in the original formulation, it becomes,
\begin{eqnarray*}
\int_{0}^{3}{\sqrt{4-x^{2}} dx} &=& \left(  3 \sqrt{1-\frac{3^{2}}{4 } }  + 2 \sin^{-1} \frac{3}{2} \right) \\
&=& \jmath \frac{3 \sqrt{5}}{2} + 2\sin^{-1} \frac{3}{2}.
\end{eqnarray*}
