Integration of $\sin(\theta)$ I hope I'm not asking a silly question. 
We can integrate $\sin(\theta)$ simply by the following identity: 
$$\int_0^\frac{\pi}{2} \sin\theta\ \mathsf d\theta = \left[-\cos\theta \vphantom{\frac 1 1} \right]_0^\frac{\pi}{2}=1.$$
But how can we do this by summation ?
For example, $$\int_0^{100} x\ \mathsf dx = \left[\frac{x^2}{2}\right]_0^{100} = 5000 \approx \sum_{x=1}^{100}x= 1+2+\cdots+100 = 5050.$$ 
How can we do the same for initially mentioned problem ?

$$\cdots\approx\frac{\pi}{ 200}\sum_{k=1}^{100}\sin\left(\frac{k}{100}\right)$$
In general:
if $b-a$ small,
$$\int_a^b f(x)dx\approx\frac{b-a}{100}\sum_{k=1}^{100}f\left(a+k\frac{(b-a)}{100}\right)$$
since $$\frac{b-a}{n}\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)\underset{n\to\infty }{\longrightarrow} \int_a^b f(x)dx.$$
