Thread: Help with Riemann sums for a continuous function

1. Help with Riemann sums for a continuous function

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.

2. I think there is a mistake. That is not expressable as a Riemann sum.

3. That's not good, I double-checked it and it is as typed in the course guide!

4. Sorry forgive me. I saw something else in the problem and made a mistake.
The function is,
$\sin \pi x$
Then, the convergent riemann sum is, (by definition)
$\int_0^1 \sin \pi x dx$

Now,
$f(x)=-\frac{1}{\pi} \cos \pi x$
Has property that,
$f'(x)=\sin \pi x$
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!)
$-\frac{1}{\pi} \cos \pi +\frac{1}{\pi} \cos 0=2\pi$

5. Rewrite the sum like so.

$\lim_{n \rightarrow \infty}\sum_{i=1}^{n} \frac{\sin(\frac{\pi{i}}{n})}{n}$

Now let $x=\frac{i}{n}$ and $dx=\frac{1}{n}$. Now all of PH's work follows.

6. The only example from my book that I have to go by is like the first image below. I tried to rewrite your work in the second image to anwser the question. Can you look at it and let me know if I messed up? Thanks

7. You must first write the sum in a general way as I did for you. You are setting up rectangles partitioned $\frac{1}{n},\frac{2}{n}$...etc and letting n approach infinity for the rectangles height, and the width of each rectangle is $\frac{1}{n}$. Now when you apply the limit, the infinite sum tranforms into the integral $\int_{0}^{1}\sin(\pi{x})dx$.