I couldn't get LaTEX to display this one, it kept erroring out after about 3/4 of the equation was entered. So I pasted a jpg instead. Thanks for looking.
Sorry forgive me. I saw something else in the problem and made a mistake.
The function is,
$\displaystyle \sin \pi x$
Then, the convergent riemann sum is, (by definition)
$\displaystyle \int_0^1 \sin \pi x dx$
Now,
$\displaystyle f(x)=-\frac{1}{\pi} \cos \pi x$
Has property that,
$\displaystyle f'(x)=\sin \pi x$
Then by the Fundamental Theorem of Calculus (eventhough I am angry when it is referred to as like this. It certainly is not a fundamental theorem at all!)
$\displaystyle -\frac{1}{\pi} \cos \pi +\frac{1}{\pi} \cos 0=2\pi$
You must first write the sum in a general way as I did for you. You are setting up rectangles partitioned $\displaystyle \frac{1}{n},\frac{2}{n}$...etc and letting n approach infinity for the rectangles height, and the width of each rectangle is $\displaystyle \frac{1}{n}$. Now when you apply the limit, the infinite sum tranforms into the integral $\displaystyle \int_{0}^{1}\sin(\pi{x})dx$.