Why is the sum of the numbers $1+2+3+4+\dots+N$ equal to the integral $\int_{0}^{N}x+0.5$, rather than $\int_{0}^{N}x$?

Question

Sorry if this question has been asked in the past, or if it seems silly, but I couldn't find anything on Google or SE, and we didn't cover series in our math classes.

So basically, I'm working on something requiring the use of a triangular kernel, so I need to find the total sum of each point sampled. To make it easier for myself, I used a single side (result multiplied by two when I'm done), so it takes the form of an easier linear equation.

I thought that to find the result of summing each point along the line y=x would be the same as taking the definite integral of 0->N to find the area under the curve (or line, in this case):

$\sum_{n=1}^{N}n = \int_{0}^{N}x\cdot dx = \frac{N^2}{2}$

However, that obviously didn't work. So then I put the sum series through Microsoft Math which gave me $\sum_{n=1}^{N}n = \frac{N^2+N}{2}$

Differentiating it, I found that the integral I was after was actually $\int_{0}^{N}(x+0.5)\cdot dx$

Why does the integral require the 0.5 offset for the sum to work the way I intended? I thought that a possible reason for this was that the integral took one number too many (ie. added N+1 to the result), but I found that wasn't the case.

Answer 1

Here is a picture for $n = 10$. The blue rectangles have width $1$ and height $0, 1, 2, \ldots, n-1$. Adding the red unit squares on top, we see the total shaded area is the sum $1 + 2 + \cdots + n$. The integral $\int_{x=0}^n x \, dx$ is the area under the line $y = x$. You can see that the two areas don't match up: The triangle doesn't include half of the area of all the red squares. But since there are $n$ red squares, their area is $n \times 1 = n$, and half of this is $n/2$. So the sum of $1 + 2 + \cdots + n$ is equal to $n^2/2 + n/2$.