Question: 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?](https://math.stackexchange.com/questions/15294/why-is-the-area-under-a-curve-the-integral/15301#15301)

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 $\pi/2$.

Answer: 
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 $\mathcal{M}(0)=0$.
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.