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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 ?

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$$\cdots\approx\frac{\pi}{ 200}\sum_{k=1}^{100}\sin\left(\frac{k}{100}\right)$$

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 $$\frac{b-a}{n}\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)\underset{n\to\infty }{\longrightarrow} \int_a^b f(x)dx.$$

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    $\begingroup$ $$f(k/100)$$ should be $$f\left(a+\dfrac{b-a}{100}k\right)$$ right? $\endgroup$ – lab bhattacharjee Aug 13 '15 at 16:56
  • $\begingroup$ thanks :-) I corrected it, $\endgroup$ – Surb Aug 13 '15 at 16:58
  • $\begingroup$ Thanks a lot for your answer. I really appreciate. Again, thanks a lot. $\endgroup$ – user3563929 Aug 13 '15 at 17:10

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