Why does the boundary of a region $D$ have enough information to dictate the value of an integral over $D$? There are many theorems which say something along the lines of the title:


*

*The FTC: $\int_a^bf'(x)dx=f(b)-f(a)$.

*Green's Theorem: Let $F=(P,Q)$, then $\oint_{\partial D}Fds=\iint_D(\frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y})dA$.

*The FTC for line integrals: If $F=\nabla g$, then $\int_{\vec p_1}^{\vec p_2}Fds=g(\vec p_2)-g(\vec p_1)$.

*The divergence theorem.

*Stoke's theorem.
Probably there are more, which I don't know.
Is there any way to get some intuition of why this happens? 
 A: Often Stokes theorem (or any of the ones that you stated) is simplified to something like "...and so we only have to look at the boundary". Which likely motivated your question. That's misleading.
The boundary does not dictate the value of the integral alone.
There's also the actual integration going on: $f' \neq f$
$$\int_a^bf'(x)dx=\overset{\color{red}{?}}{f}(b)-\overset{\color{red}{?}}{f}(a)$$
No, if you know $f'$, $a$ and $b$ you do not know the value of the integral , because you need to know $f$ to take advantage of the equation above. It's true that the calculation of the value of the integral is easy given that you know the integrated function, but in general you don't know the integrated function.
intuitive explanation
The key concept of why these theorems work intuitively is to think about them in terms of rates of change. The derivative of some variable is a value that describes how much that variable changes. Speed is a common example. Speed is the change of position.
Let's say $f$ is the position of your computer mouse and thus $f'$ is its speed. (Which are both one dimensional for the sake of this example) Let's change the above formula:
$$\int_{morning}^{evening}\text{speed}_\text{mouse}(t)dt=\text{position}_\text{mouse}(evening)-\text{position}_\text{mouse}(morning)$$
You probably moved your computer mouse back and forth today. If you wanted to know what difference this day made on the position of your mouse, you could remember the position it had in the morning and compare it to the position in the evening (the right side of the equation):
$$\text{position}_\text{mouse}(evening)-\text{position}_\text{mouse}(morning)$$
If you think about it in rates of change instead, you could say that at all times, your mouse had a certain $\text{speed}_\text{mouse}$ (which represents how the $\text{position}_\text{mouse}$ changes) and if you look at the entirety of all changes made to a position over the day (the left side of the equation)
$$\int_{morning}^{evening}\text{speed}_\text{mouse}(t)dt$$
they should sum op to the overall difference in position as denoted by the previous equation.
You can look at the value from either side:


*

*What's the difference between the value at the beginning and the end?

*What's the sum of all changes made to the value between the beginning and the end?


and the result is the same.
The benefit of using the theorems in the example: It's easier to measure the position of your computer mouse at two points in time per day than to measure its speed continuously. It's true that only dealing with the boundary is beneficial here, but it requires you to be able to know the $\text{position}_\text{mouse}$. You are dealing with different kind of value (the integral) and as a result, you only need to deal with the boundaries.
