Why does $\int^{b}_{a} f(x)dx-\int^{b}_{a}g(x)dx = \int^{b}_{a} [f(x)-g(x)]dx$ for area between curves.

Question

How does $f(x)$ and $g(x)$ combine in the next line to become $[f(x)-g(x)]$? I would like to know algebraically and if there is a graphical link between finding area between the curves. For example from 0 to 1 for $y=x^2$ and $y=x$, what does $x-x^2$ have to do with it? Why is the area between the curves for the first two functions equal to the area above the x-axis of $y=x-x^2$?

Answer 1

The following is always true:

1. For any two integrable funtions $f_1, f_2$, that $$\int_a^b (f_1(x) + f_2(x)) dx = \int_a^b f_1(x)dx + \int_a^b f_2(x)dx$$
2. For any scalar $\alpha$ and any integrable function $h$, that $$\int_a^b \alpha \cdot h(x)dx = \alpha \cdot \int_a^b h(x) dx$$

These two rules are enough to show that $\int^{b}_{a} f(x)dx-\int^{b}_{a}g(x)dx = \int^{b}_{a} [f(x)-g(x)]dx$

Now, additionally knowing that $\int^{b}_{a} f(x)dx$ is the area under $f$, and $\int^{b}_{a} g(x)dx$ is the area under $g$, subtracting the two means you want the area under $f$ that is not under $g$, and another word for that is the area between $f$ and $g$.