Question: $$\int\_{-π/2}^{π/2} \cos t \ dt = 2$$

Though I understand it mechanically, the idea of $F(b) - F(a)$. Conceptually, I'm missing something. If you draw $\sin{x}$ for $-\pi/2$ to $\pi/2$ in the unit circle you end up with half a circle. Since there's a bit of area between the arc and the straight lines connecting $\pi/2$ to $\pi$ and $\pi$ to $-\pi/2$, I wonder what happens with this area.

Answer: 
Continuity of the partial derivatives is strong sufficient condition. 

Consider the simplest proof of Green's theorem for a rectangular region $D = [a,b] \times [c,d]$. One step is to show that (with $P\_y := \frac{\partial P}{\partial y}$)

$$\tag{\*}\oint\_C P(x,y) \, dx = -\int\_D P\_y(x,y) \, dA,$$

which reduces to 

$$\int\_a^b P(x,c) \, dx - \int\_a^b P(x,d) \, dx = -\int\_a^b \int\_c^d P\_y(x,y) \, dx \, dy = -\int\_a^b \left(\int\_c^d P\_y(x,y) \, dy\right) \, dx$$ 

We have to be sure that we can evaluate the double integral as an iterated single integral in any order and we have to apply the fundamental theorem of calculus to obtain

$$\int\_c^d P\_y(x,y) \, dy = P(x,d) - P(x,c),$$

in showing that (\*) is true.

All of these steps can be justified if the partial derivatives are continuous.

However, there is a more general form of Green's theorem for Lebesgue integrals which requires only that $P$ and $Q$ be absolutely continuous. In this case, the partial derivatives exist almost everywhere, are integrable, and the fundamental theorem still holds even though the derivatives need not be continuous everywhere.

For example, observe that Green's theorem still holds for $D = [0,1] \times [0,1]$ and

$$P(x,y) = 0 \\ Q(x,y) = \begin{cases}yx^2\sin(1/x), \, \, x \neq 0 \\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x = 0 \end{cases}$$

even though $\frac{\partial Q}{\partial x}$ is not continuous on $\{(x,y): x = 0, 0 < y \leqslant 1 \}$.
An example you're looking for, where the partial derivatives, $\frac{\partial P}{\partial y}$, $\frac{\partial Q}{\partial x}$ are discontinuous is when $P = \frac{-y}{x^2 + y^2}$ and $Q = \frac{x}{x^2+y^2}$. 

This yields $\frac{\partial P}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2}$, $\frac{\partial Q}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2}$, which are discontinuous at $(x,y) = (0,0)$. If we parametrize $x$, and $y$, by $$\begin{cases}
 x = cos\theta \\
 y = sin\theta \\ 
 0 \leq \theta \leq 2\pi 
\end{cases} \implies \begin{cases}
 dx = -sin\theta\ d\theta \\
 dy = cos\theta\ d\theta, 
\end{cases}$$

and calculate the line integral over the unit disc, $$\int\_{|r| = 1}^{} \frac{-y\ dx}{x^2+y^2} + \frac{x\ dy}{x^2+y^2} = \int\_{|r| = 1}^{} \frac{(-sin\ \theta)^2\ d\theta}{1} + \frac{(cos\ \theta)^2\ d\theta}{1} = \int\_{|r| = 1}^{} d\theta = 2\pi.$$

However, calculating $\int\_{|r| = 1}^{} \frac{-y\ dx}{x^2+y^2} + \frac{x\ dy}{x^2+y^2}$ by applying Green's Theorem, we obtain a contradiction, $$\int\_{|r| = 1}^{} \frac{-y\ dx}{x^2+y^2} + \frac{x\ dy}{x^2+y^2} = \iint\_{|r| \leq 1}(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dxdy = \iint\_{|r| \leq 1} 0\ dxdy = 0 \neq 2\pi.$$

Since all conditions of Green's theorem are met, except for the continuity of the partial derivatives of $P$, and $Q$ at the origin, combined with the fact that we arrive at a contradiction, we conclude that the partial derivatives of $P$, and $Q$, has to be continuous for all values of the unit disc. 

Another way to interpret this is by trying to use Green's Theorem on the punctured disc with the origin removed. On this new domain the partial derivatives of $P$, and $Q$, are continuous, but in this case we do not obtain a region $D$ bounded by the corresponding positively oriented, piecewise-smooth, simple closed curve in the plane, i.e the unit circe in the example above.