$1+2+3+\cdots+x > \text{Integral of } x$ Why is $\Sigma_{n=1}^xn$, or $\frac{x^2+x}{2}$,
bigger than $\int x\ \text dx$, even though the integral of $x$ adds up more numbers than $\frac{x^2+x}{2}$ ?
 A: This result is apparent from looking at a graph. In the following (rough) picutre, you can see that the blue rectangles have areas $1$ and $2$. The total of their areas is thus $1+2$, while the integral $\int_0^2 x \,dx$ is only the area under the black line, which is less than the area of the combined rectangles.
The sum $1+2+\cdots+n$ is actually equal to the integral $\int_0^n \lceil x\rceil \,dx$, where $\lceil x\rceil$ is the ceiling function, which satisfies $\lceil x\rceil \geq x$ for all $x$.
The sum "adds up" $1$ all the way from $0$ to $1$, while the integral "adds up" mostly numbers smaller than $1$ over that same interval.

A: For one thing, it isn't really meaningful to talk about "how many numbers" the integral is adding. We can think of it as the limit of a sum of increasingly many terms, but the terms are getting smaller and smaller.
But to understand why the sum is bigger than the integral, draw a picture of the two. The sum looks like a staircase, while the integral is a straight line that cuts the corners off the staircase.
A: As @Theophile said, the sum $1+2+3+\cdots+x$ are the rectangles like a 'staircase', while the $x$ is like a smooth slope, as shown in this graph from Desmos.
The integral of $x$, which is $\frac{x^2}{2}$ approximates the slope by adding up infinitely many rectangular 'strips' of infinitely small length. As the number of 'strips' increases, the 'strips' better approximate the smooth slope of $y = x$. Therefore, when we add infinitely many 'strips', the area of $x$ will be exactly the same as $1+2+3+\cdots+$.
However, in this case, the rectangles have a width of $1$. Since the width of the 'strips' is not infinitely small, there will be some extra area between the two functions. 
