How integration arises from summation

Question

I'm trying to understand how integration arose out of Leibniz's work on sequences and series. I've asked over at HSM, but the substance of the question is more maths than history of maths.

I understand that integration began as a form of summation. Indeed, $\int$ and $\sum$ are both 'S' for 'sum' or 'summation'. Therefore I expect that the sum of a sequence and the integral of the corresponding algebraic expression should agree. But consider the following.

Given $y=x$:

from Cavalieri's Quadrature Formula it follows that $$\int y\;dx = \frac{x^2}{2}+c$$

but by summation (I used trial and error) it appears that

$$\sum_{y=0}^x y = \frac{x^2+x}{2}$$

Why don't they agree?

Edited for clarity.

Second edit: With some prompting from the answerers I think I've got the answer I was looking for. The summation is an upper bound of the integral. A corresponding lower bound would be given by $\frac{x^2-x}{2}$. Then the familiar process of letting x go to 0 brings the bounds together, producing the integral.

Answer 1

Here I am trying to show why the summation expression and the integration expression are different.

$$\sum_{y=1}^x y = \frac{x^2+x}{2}$$ The above summation produces an equation to calculate a value representing the sum of integers $(1,2,...,x)$.

The following integration produces a function, an anti-derivative, since it is not a bounded integral. Given $y=f(x)=x$: $$\int x\;dx = \frac{x^2}{2}+c$$

Note that $c$ is not determined (and can't be determined from just this information).

The above expression tells that: $$\frac{d}{dx}\left(\frac{x^2}{2}+c\right)=x$$

At this point, we can see why the two expressions don't yield the same result. They address different concepts completely.

You can use bounded-integral to calculate the area under a curve (or intersection area of curves). In this case you get a constant function that represents the value of the area under the function curve in arbitrary square units.

As an example consider:

$$\int _1^3\:x\:dx=4$$

The concept of summation and bounded integral are related. Area under the curve could be computed approximately if you divide the curve into rectangles, with an equal base. In this case the higth would be $f(x)$.

The approximate area would be (with base=1, and height $f(x)=x$)

$$\sum _{x=1}^{3\:}f(x).base=\sum _{x=1}^{3\:}x.1=6$$

The obtained area using this method is a bad approximation of the area. This is because the base of 1 is so large and you can see that part of the rectangle will be above $f(x)$ for all values of $x$. The value of the bound integral tends to come closer to the summation value as you make the base infinitely small, hence getting very large of vertical rectangles.

A more general view:

A better method to approximate the area following a close concept is to use the Trapezoidal Rule. There are much better methods still.

Figure Reference: Introduction to Integration