Integral of $\sqrt{4-x^2}$ by Riemann sum I was trying to compute $\int_{0}^{2}\sqrt{4-x^2}\,dx$ by Riemann sum like this
$$\sum_{i=1}^{n}\sqrt{4-x_i^2}\,\Delta x_i$$
where $x_i = \frac{2i}{n}$ and $\Delta x_i = \frac{2}{n}$, but I couldn't get something useful for the answer. I think it should be with another kind of partition, but I don't know which kind of partition to use.
 A: The right Riemann sum using the partition with points $x_k = 2 \sin \frac{\pi k}{2n}\,\,(k = 0,1,2,\ldots, n)$ is
$$\tag{*}S_n = \sum_{k=1}^n \sqrt{4 - 4  \sin^2 \frac{\pi k}{2n}}\left(2  \sin \frac{\pi k}{2n}- 2\sin \frac{\pi (k-1)}{2n}\right)\\ = 4\sum_{k=1}^n \cos \frac{\pi k}{2n}\left(  \sin \frac{\pi k}{2n}- \sin \frac{\pi k}{2n}\cos \frac{\pi}{2n}+ \sin \frac{\pi }{2n}\cos \frac{\pi k}{2n}\right)\\ = 4\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n} + 4 \sin \frac{\pi}{2n}\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}$$
Note that
$$\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n}= \frac{1}{2}\sum_{k=1}^n\sin \frac{2\pi k}{2n} =\frac{1}{2}\sum_{k=1}^n\sin \frac{\pi k}{n},\\\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}= \sum_{k=1}^n \left(\frac{1}{2} - \frac{1}{2} \cos \frac{2\pi k}{2n} \right) = \frac{n}{2}- \frac{1}{2}\sum_{k=1}^n\cos \frac{\pi k}{n}$$
Substituting into (*), we get
$$S_n = \underbrace{2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}}_{A_n} + \underbrace{2n \sin \frac{\pi}{2n}}_{B_n}- \underbrace{\sin \frac{\pi}{2n}\sum_{k=1}^n \cos\frac{\pi k}{n}}_{C_n} $$
It is shown below that $A_n - C_n \to 0$ as $n \to \infty$, and, therefore,
$$\int_0^2 \sqrt{4-x^2}\, dx= \lim_{n \to \infty}S_n = \lim_{n \to \infty}2n \sin \frac{\pi}{2n} = \lim_{n \to \infty}\pi \frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}} = \pi$$

Limits of $A_n$ and $C_n$:
We have, since $\frac{1-\cos x}{\sin x} \to 0$ as $x \to 0$,
$$A_n = 2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}= 2 \left(1 -\cos\frac{\pi}{2n} \right) \frac{\sin \frac{n}{2}\frac{\pi}{n}\sin \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = 2 \frac{1 -\cos\frac{\pi}{2n}  }{\sin \frac{\pi}{2n}}\sin \left[\left(1 + \frac{1}{n}\right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
Also,
$$C_n = \sin\frac{\pi}{2n} \sum_{k=1}^n \cos \frac{\pi k}{n} =  \sin\frac{\pi}{2n}\frac{\sin \frac{n}{2}\frac{\pi}{n}\cos \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = \cos \left[\left(1+\frac{1}{n} \right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
A: 
Note that the partition you mentioned is not in fact a partition of the interval $[0, 2].$ I think that what you would want to do involves considering the partition $P = (\xi_i)_{1 \leq i \leq n},$ where $\xi_i := \frac{2i}{n}$ for all $1 \leq i \leq n$ (and, of course, $\Delta \xi_i = \frac{2}{n}$).
