Converting a Riemann sum to a definite integral.

So, I'm supposed to find $$\lim_{n \to \infty} \sum_{i=1}^n \frac{sin(i \pi /n)}{n}$$ by evaluating an appropriate definite integral over the interval [0, 1]. Here's my work so far: $$\lim_{n \to \infty} \sum_{i=1}^n \frac{sin(i \pi /n)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \left[ \sin \left(\frac{i \pi}{n} \right) \cdot \frac{1}{n}\right]$$
$$= \int_0^1\! sin(x) dx$$
$$= 1 - cos(1)$$
No clue if I'm on the right track here, any help is appreciated.

Question

Converting a Riemann sum to a definite integral.

So, I'm supposed to find $$\lim_{n \to \infty} \sum_{i=1}^n \frac{sin(i \pi /n)}{n}$$ by evaluating an appropriate definite integral over the interval [0, 1]. Here's my work so far: $$\lim_{n \to \infty} \sum_{i=1}^n \frac{sin(i \pi /n)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \left[  \sin \left(\frac{i \pi}{n} \right) \cdot \frac{1}{n}\right]$$
$$= \int_0^1\! sin(x) dx$$
$$= 1 - cos(1)$$
No clue if I'm on the right track here, any help is appreciated.

ganeshie8 · Answer

integrand should be $\large  \sin(\color{red}{\pi} x)$ right ?

anonymous · Answer

Is that what I missed? I was wondering where to put that $ \pi$ symbol.

ganeshie8 · Answer

Yes only that, everything else looks perfect!

anonymous · Answer

Sweet! Thank you, Ganeshie8 :)

ganeshie8 · Answer

you're welcome :)

ganeshie8 · Answer

wolfram is your best friend in double checking the final answer http://www.wolframalpha.com/input/?i=lim+n%5Cto+%5Cinfty+%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7Dsin%28ipi%2Fn%29%2Fn+

anonymous · Answer

Ah sweet, I tried to use wolframalpha to check but I couldn't figure out how to input the expression; I didn't know you could basically input the LaTeX code.

ganeshie8 · Answer

yes copy pasting the latex code into wolfram works almost all the time !

anonymous · Answer

Sweet! that'll help with checking a lot of my answers, thank you.