Geometric meaning of the formula Suppose $f(x)$, $g(x)$ to be the functions and $[a,b]$ the interval. Then
$$\int_{a}^{b}\left(f(x)-g(x)\right)dx$$
represents the area "between" $f(x)$, $g(x)$ on $a,b$. However, what is the "geometric" meaning of the squared difference
$$\int_{a}^{b}\left(f(x)-g(x)\right)^2dx$$
of $f(x)$, $g(x)$?
Thanks for your help.
 A: One geometric meaning of $\int_a^b (f(x)-g(x))^2\,dx$ is the square of the distance between the functions $f$ and $g$. When I say "distance" I mean distance in the sense of a normed vector space. That is, you can make a (huge!) space whose vectors are ($L^2$-integrable) functions from $[a,b]$ to $\mathbb R$ and where you measure the size of a vector (i.e. function) $f$, written $\|f\|$, by
$$
\|f\|=\bigg[\int_a^b f(x)^2\,dx\bigg]^{1/2}.
$$
This size function has the properties that 


*

*$\|f\|=0\iff f=0$ (The only function with zero size is $0$. This should be obvious.)

*$\|af\|=|a|\|f\|$ for $a\in\mathbb R$ (Size respects scaling. This should be obvious.)

*$\|f\|+\|g\|\geq\|f+g\|$ (The triangle inequality. This is probably not obvious.)


and these properties are what allow me to say that it "measures size".
Then whenever you have two functions $f$ and $g$, the "size" of $f-g$ is simply the distance between them, and
$$
\|f-g\|^2=\int_a^b(f(x)-g(x))^2\,dx,
$$
as I claimed at the beginning of this answer.
I used a lot of terms that you may not be familiar with. Did that make any sense?
