You are looking for the area under the curve $f(x) = - \sin x$ between the values $x = -\pi/4$ and $x = \pi/6$.
$$\int_{-\pi/4}^{\pi/6} -\sin x \,dx$$
But if you're looking for area between the curve $f(x) = -\sin x$ and the $x$-axis, then you need to divide the integral into two integrals if you are to find "absolute area" between the curve and the x-axis, which is the curve/line $y = 0$, since on the interval $(0, \pi/6)$, $-\sin x < 0$.
Close up of positive region (above x-axis) and negative region (below x-axis).
So if you want area between the curve and the x-axis (i.e., the line $y = 0$), you need to integrate the following two integrals, and in each, we take the top curve and subtract the bottom curve. For the first integral, $-\sin x > 0$ so the integrand is $-\sin x - 0$, and in the second integral, $0 > -\sin x$, so we need for the integrand to be $0 - (-\sin x)$:
$$\begin{align}\text{Area}\;& =\;\int_{-\pi/4}^0 (-\sin x - 0)\,dx + \int_0^{\pi/6}(0 - (-\sin x))\,dx \\ \\
& = \int_{-\pi/4}^0 -\sin x\,dx \quad +\quad \int_0^{\pi/6}\sin x\,dx \\ \\
& = \cos x\Big|_{-\pi/4}^0 \quad + \quad -\cos x\Big|_0^{\pi/6}\\ \\
& = \left(1 - \frac{\sqrt 2}2\right) - \left(\frac{\sqrt 3}2 - 1\right)\\ \\
& = 2 - \frac 12(\sqrt 3 + \sqrt 2) \end{align}$$
This stems from one of the "fundamental" theorems of calculus. You are asking, I think, why is it the case that whenever $F'=f$, it follows that we can compute
$$\tag{1}F(b) -F(a) = \int_a^b f(x)dx.$$
The idea is as follows. Define $G(t) = \int_a^t f(x)dx$, so this gives "the area up to a certain time $t$". Then, naturally, you are looking for $G(b)$, but (unfortunately) we do not know how to compute this.
The insight of equation $(1)$ is that the function $G$ can be differentiated, and that its derivative equals $f$. To see why this is the case, we note that
$$ G(t+h)-G(t) = \int_t^{t+h} f(x)dx = h\cdot f(\xi)$$
for some mid-point $\xi\in [t,t+h]$. As $h\to 0$, we see that $\xi\to t$, and assuming $f$ is continuous (which is the case, at least usually in first calculus courses) we get that
$$G'(t) = f(t).$$
The takeaway is that, because any two functions $F$ and $G$ with $F' = G'$ differ by a constant, we see that $F(t) - G(t)$ is constant, and this means that
$$F(b)-G(b) = F(a) - G(a)$$
and a little rearranging (plus $G(a)=0$) shows
$$F(b) - F(a) = \int_a^b f(x)dx.$$
Best Answer
[image source]
You can think of the integral evaluated at a certain point as the area under the function's curve up until that point (where we call the area negative if the function is progressing below the $x$-axis and positive if it is progressing above the $x$-axis). So the integral evaluated at $b$ is the total area (under these sign conventions) under the curve up until $x=b$, and the integral evaluated at $a$ is the total area under the curve up until $x=a$. You can see intuitively from looking at the picture why subtracting the latter from the former would give us the shaded area. This is certainly not rigorous, but it may be the geometric intuition you are looking for.