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 $\pi/2$.
 A: In the notation of the post you linked to, you are confusing $\mathcal{M}$ and $F$.  What you are calling $F(b)$, a function which measures the area under the curve from $0$ to $b$, is what the post calls $\mathcal{M}(b)$.  The function $F$ is instead any antiderivative of $f$.  That is, we know nothing about $F$ at all other than that it is some function whose derivative is $f$.
This doesn't mean that $F$ is the same as $\mathcal{M}$, since a function can have more than one antiderivative!  Indeed, for any constant $C$, $F(x)=\mathcal{M}(x)+C$ is another antiderivative of $f$, since adding a constant does not change the derivative.  However, this is the only way to get another antiderivative: if $F$ is an antiderivative of $f$, then $F'(x)-\mathcal{M}'(x)=f(x)-f(x)=0$, so the function $F(x)-\mathcal{M}(x)$ has derivative $0$ and hence is a constant.
So, we're starting with some function $F$ which is an antiderivative, but what we actually want is $\mathcal{M}$.  To fix this, we have to subtract a constant from $F$, and that constant is exactly $F(0)$, since $\mathcal{M}(0)=0$.
A: Adding to what Dylan said, I think you are misunderstanding how to calculate a definite integral, as in your example: There are three steps to calculating one, which are:
$1)$ Find the indefinite integral - the integral without the bounds:
$\int \sin x = -\cos x$ (the classical way is to prove this by Taylor series, not sure about any others)
$2)$ Find the bounds: The upper bound is $-\cos{\pi/2}$ which is $0$, the lower bound is $-\cos{0}$, which is $-1.$
$3)$ Subtract the lower bound from the upper bound: $0 - (-1)$ equals $1$, which is the signed area - the area above the $x$-axis minus the area below. 
A: 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.

A: 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?
