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 $\pi/2$.

From my point.of view a simple way to see this fact is to consider the integral function:
$$F(x) = \int_{0}^{x}f(t)dt $$
that rapresent the area “under” the graph from 0 to x.
Now if we think to calculate its derivative is pretty clear that for a small change $\Delta x$ the area varies of the quantity:
$$\Delta F(x)=f(x)\cdot \Delta x$$
Thus the rate of change is
$$\frac{\Delta F(x)}{\Delta x}=f(x) $$
and in the limit
$$F’(x)=f(x)$$
That’s the link between the two concept.

Now, since the derivative of a constant is zero, any constant may be added to an antiderivative and will still correspond to the same integral (i.e. the antiderivative is a nonunique inverse of the derivative).
For this reason, indefinite integrals are written in the form
$$\int f(x)dx = F(x) + c$$
where $c$ is an arbitrary constant of integration.
For this reason when we evaluate a definite integral we need to calculate it as a difference, just to eliminate the constant of integration.

