Why does the difference $F(b)-F(a)$ give an area below a curved line? So I understand how to calculate integrals and that it's the area under the curve. What I'm struggling to understand is how two points subtracted from one another give the area under a curved line.
For example, if I have a slope function $y=f(x)$, then to find the area under the curve between two points  I need to evaluate the definite integral. Why doesn't this definite integral just give me a linear line between the two points that I'm evaluating? How does calcuating the difference between two points give me everything below a curved line?
 A: This stems from one of the "fundamental" theorems of calculus. You are asking, I think, why is it the case that whenever $F'=f$, it follows that we can compute
$$\tag{1}F(b) -F(a) = \int_a^b f(x)dx.$$
The idea is as follows. Define $G(t) = \int_a^t f(x)dx$, so this gives "the area up to a certain time $t$". Then, naturally, you are looking for $G(b)$, but (unfortunately) we do not know how to compute this.
The insight of equation $(1)$ is that the function $G$ can be differentiated, and that its derivative equals $f$. To see why this is the case, we note that
$$ G(t+h)-G(t) = \int_t^{t+h} f(x)dx = h\cdot f(\xi)$$
for some mid-point $\xi\in [t,t+h]$. As $h\to 0$, we see that $\xi\to t$, and assuming $f$ is continuous (which is the case, at least usually in first calculus courses) we get that
$$G'(t) = f(t).$$
The takeaway is that, because any two functions $F$ and $G$ with $F' = G'$ differ by a constant, we see that $F(t) - G(t)$ is constant, and this means that
$$F(b)-G(b) = F(a) - G(a)$$
and a little rearranging (plus $G(a)=0$) shows
$$F(b) - F(a) = \int_a^b f(x)dx.$$
A: if we have a function $f(x)$ which is continuous over some domain $x\in[a,b]$ then in the same way that we can take derivates and define these as functions, we can think of indefinite integrals as the negativith derivative (e.g. $f^{(-1)}(x)=F(x))$ Now trying to work from a function to its integral and seeing the area can be difficult and is best explained using the [Riemann integral definition][1] to see what the integral operation actually does.
One of the easiest ways to think about it is in reverse, say you have a function $F(x)$ which satisfies $F'(x)=f(x)$. Once you understand this concept you can start to understand how the fundamental theorem of calculus is applied. We can define:
$$F(x)=\int f(x)dx+C\tag{1}$$
and now for the ease of notation, definite integrals are written as $F(b)-F(a)$, in other words the difference between the "area" of a function in said given domain. As you can see when plugging this into the above formula, the $+C$ and $-C$ will cancel out, which is why constants of integration are irrelevant in single definite integrals.
[1]: https://en.wikipedia.org/wiki/Riemann_integral#:~:text=Loosely%20speaking%2C%20the%20Riemann%20integral,more%20specifically%20Riemann%2Dintegrable).
