Dividing given interval into n equal parts and for every interval taking ϵ as an midpoint of an interval find integral sum of a function and compute it.
$y=\sin x$ $x \in[0,\pi]$
$\triangle x_i= \frac{\pi}{n}i$
$\epsilon_i= \frac{0+\frac{\pi}{n}i+0+\frac{\pi}{n}(i+1)}{2} = \frac{2i+1}{2n}\pi$
$$\sum_{i=0}^{n-1}\sin\left(\frac{(2i+1)\pi}{2n}\right)\cdot\frac{\pi}{n} = \frac{\pi}{n}\sum_{i=0}^{n-1}\sin\left(\frac{(2i+1)\pi}{2n}\right)$$ from here I can't continue