Why $\int_a^bf(x)\,dx=F(b)-F(a)$ I recently looked at the proof given for the fundamental theorem of calculus in this link:
Why is the area under a curve the integral?
It made perfect sense, except for one thing. The proof relies on creating a function $F(b)$ that gives the area under the curve $f(x)$ from $0$ to some real number $b$ so that in essence $F(b) = \int_{0}^{b}f(x)dx $ and then proved that $F(b)$ is the anti derivative of $f(x)$. 
However, if we define the integral in this way, then it seems strange that when integrating a function from 0 to a value b we have to evaluate $F(b)-F(0)$ rather than just evaluate $F(b)$. Since the former would generally imply that if we want to find the area under the curve from a to b, then given the definition of an integral, we simply have to subtract the area from 0 to a from the area from 0 to b. Which in this case makes no sense, since we would be subtracting the area from 0 to 0, ie. 0 from the area from 0 to b. Which means we could just discard the first part of the evaluation, yet this would cause us problems if we wanted to evaluate something like $ \int_{0}^{\pi/2}sin(x)dx $ which would be zero if we just evaluate the antiderivative of sin(x) at $\pi/2$.

An antiderivative of $f$ is a function that collects area under the graph of $f$, starting somewhere. In order to find the area from $a$ to $b$, we need to plug $x=b$ into the particular antiderivative that collects area starting at $a$. One name for this function is $$\int_a^x f(t) \,dt$$
When we write down some random antiderivative $F$, though, we don't know if it starts in the right place. However, if we subtract $F(a)$, then we have $F(x)-F(a)$: an antiderivative that has a value of $0$ at $x=a$, which is perfect.
The integral from $a$ to $b$ is what we get when we plug $b$ into the specific antiderivative given by $F(x)-F(a)$.
Does this help?
