# Riemann sums with limits

I just learned about approximation using Riemann Sums, and all that has been taught to us was how to approximate the area under the curve using rectangles. Now, I wanted to try my hand at generalizing this, and came up with the following approach (I'm pretty sure it's horrifyingly naive).

$$\lim_{\Delta x\rightarrow0}\Delta x\left(\sum_{i=0}^{\frac{x_1 - x_2}{\Delta x} - 1}f(x_1 + \Delta xi)\right)$$

Where $x_1$ and $x_2$ are the upper and lower boundaries respectively.

I'm pretty sure there's a much better method for the area under the curve. But just out of curiosity, what will the above evaluate to?

What I basically did was, I made the width of each rectangle approach zero. Now, my intuition says that $\Delta x$ will approach zero, while the sum approaches $\infty$. But, that just doesn't make sense. Will the above be undefined?

EDIT: Assume that we are flooring the upper limit of the sum.

• The formulation is problematic because the upper limit for $i$ is not necessarily an integer. – Thomas Andrews Jul 7 '13 at 15:26
• $0 \cdot \infty$ is an indeterminate form, i.e. the theorem on the product of limits doesn't work in this case but the limit may exist finite or infinite apart from this. – Tony Piccolo Jul 8 '13 at 6:44

If $f\colon[x_1,x_2]\to\mathbb R$ is Riemann integrable (and especially bounded near $x_2$), the limit is the Riemann integral.