Come on! Some of us use integrals, or even the
fundamental idiom of integral calculus
to explain the idea that areas can be subtracted. However,
the concept of subtracting areas is much more general, and much older, than this.
An area is an area; integrals are just a means to calculate them; more often they aren't even used that way.
If I'm not mistaken, subtraction of areas goes all the way back to Euclid.
As an example. Starting with a rectangle (with even more rectangles in it), we can do the following to deduce
the area of an arbitrary triangle. Herewith it is only assumed that the formula for the area of a rectangle
is well understood.
To begin with, a parallelogram $\overline{ABDC}$ is constructed inside a rectangle $\overline{AEDF}$:
Label the vertices and assign a few coordinates (addition of vectors giving $D$):
$$ A = (0,0) \quad , \quad B = (x_1,y_1) \quad , \quad C = (x_2,y_2) \quad \Longrightarrow \quad D = (x_1+x_2,y_1+y_2)$$
Then calculate areas according to the following, leaving it to the user to prove things where they feel the need. Though,
with Euclidean geometry, What you see is what you get, most of the time:
$$
\operatorname{Area}(\overline{CDf}) = \operatorname{Area}(\overline{AbB}) =
\frac{1}{2} \operatorname{Area}(\overline{AbBh}) = \frac{1}{2} \overline{Ab}\times\overline{Ah} = \frac{1}{2} x_1y_1 \\
\operatorname{Area}(\overline{BcD}) = \operatorname{Area}(\overline{ACg}) =
\frac{1}{2} \operatorname{Area}(\overline{AaCg}) = \frac{1}{2} \overline{Aa}\times\overline{Ag} = \frac{1}{2} x_2y_2 \\
\operatorname{Area}(\overline{bEcB}) = \operatorname{Area}(\overline{gCfF}) = \operatorname{Area}(\overline{Aakh}) =
\overline{Aa}\times\overline{Ah} = x_2y_1
$$
Now subtract the areas of $\overline{AbB} , \overline{bEcB} , \overline{BcD} , \overline{CDf} , \overline{gCfF} ,
\overline{ACg}$ from the area of the big rectangle,
which is $\operatorname{Area}(\overline{AEDF}) = (x_1+x_2)(y_1+y_2)$. Then what we get is the area of the parallelogram:
$$
\operatorname{Area}(\overline{ABDC}) = (x_1+x_2)(y_1+y_2) - 2 x_2y_1 - 2\cdot\frac{1}{2} x_2y_2 - 2\cdot\frac{1}{2} x_1y_1 =\\
x_1y_1+x_1y_2+x_2y_1+x_2y_2 - x_2y_2 - x_1y_1 - 2x_2y_1 = x_1y_2-x_2y_1
$$
Do we observe a determinant here? Ayway, the area of the triangle is half of this:
$$
\operatorname{Area}(\overline{ABC}) = \frac{1}{2}(x_1y_2-x_2y_1) = \frac{1}{2} \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix}
$$
There are many interesting applications. For example, the area of an arbitrary polygon can be calculated with addition and
subtraction of nothing else but triangles, as exemplified here.
However, there is a tag (definite-integrals) associated with the question, so let's kill the mosquito with a gun end employ some of that in the end:
$$
\operatorname{Area}(\overline{ABC}) = \operatorname{Area}(\overline{AaC}) + \operatorname{Area}(\overline{abBC})
- \operatorname{Area}(\overline{AbB}) =\\ \int_0^{x_2} \frac{y_2}{x_2}x\,dx \;
+ \; \int_{x_2}^{x_1} \left[y_2 + (y_1-y_2)\frac{x-x_2}{x_1-x_2} \right] dx \; - \; \int_0^{x_1} \frac{y_1}{x_1}x\,dx = \\
\frac{1}{2}x_2y_2 + \frac{1}{2}(x_1-x_2)(y_2+y_1)-\frac{1}{2}x_1y_1 = \frac{1}{2}(x_1y_2-x_2y_1)
$$
Always nice to see the consistency of mathematics showing up again.
