Show that $\int_a^b f(u) \int_a^u f(v) dv du=\int_a^b f(u) \int_u^b f(v) dv du$ Let $f:[a,b]\rightarrow \mathbb{R}$ be integrable on $[a,b]$. Show that ($\forall u\in[a,b]$)
$$\int_a^b f(u)\int_a^u f(v) dv\; du=\int_a^b f(u)\int_u^b f(v) dv\; du$$
What i've done so far: I suppose $\int f$ has $F$ as an antiderivative (which is not guaranted by $f$ being just integrable, but i can't think other way). So, 
$$\int_a^b f(u)\int_a^u f(v) dv\; du=\int_a^b F(u)f(u)\;du-F(a)\int_a^b f(u) du=\int_a^b F(u)f(u)\;du+F(a)^2-F(a)F(b)$$
$$\int_a^b f(u)\int_u^b f(v) dv\; du=F(b)\int_a^b f(u)\;du-\int_a^b F(u)f(u) du=
F(b)^2-F(a)F(b)-\int_a^b F(u)f(u)\;du$$
In terms of areas, for let´s say $f(x)=2x$ with $[-2,3]$, and $F(x)=x^2$ what we're integrating from $-2$ to $3$ is: $$(F(u)-F(a))f(u)=
2(u^2-(-2)^2)u=2(u^2-4)u$$ and
$$(F(b)-F(u))f(u)=2(9-u^2)u$$
If i graph these from $-2$ to $3$, they don't seem related, but the area under both curves is the same.
Can someone give me an approach. ¿Is this a known result? Thanks 
 A: For $ u\in\left[a,b\right] $ define a function $ f_{u}:\left[a,b\right]\rightarrow\mathbb{R} $ as following : $ f_{u}\left(v\right)=\left\lbrace\begin{aligned}f\left(u\right)f\left(v\right),\ \ \ \ &\textrm{If }a\leq v\leq u\\ 0\ \ \ \ \ \ \ \ ,\ \  \ \ &\textrm{If }u\leq v\leq b\end{aligned}\right. \cdot $
Fubini's theorem allows us to write the following : $$ \int_{a}^{b}{\int_{a}^{b}{f_{u}\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{\int_{a}^{b}{f_{u}\left(v\right)\mathrm{d}u}\,\mathrm{d}v} $$
Since the left side can be written as : $ \int\limits_{a}^{b}{\int\limits_{a}^{b}{f_{u}\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int\limits_{a}^{b}{\int\limits_{a}^{u}{f\left(u\right)f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int\limits_{a}^{b}{f\left(u\right)\int\limits_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u} \cdot $
And the right side can be written as : $ \int\limits_{a}^{b}{\int\limits_{a}^{b}{f_{u}\left(v\right)\mathrm{d}u}\,\mathrm{d}v}=\int\limits_{a}^{b}{\int\limits_{v}^{b}{f\left(u\right)f\left(v\right)\mathrm{d}u}\,\mathrm{d}v}=\int\limits_{a}^{b}{f\left(v\right)\int\limits_{v}^{b}{f\left(u\right)\mathrm{d}u}\,\mathrm{d}v} \cdot $
We have : $$ \int_{a}^{b}{f\left(u\right)\int_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{f\left(v\right)\int_{v}^{b}{f\left(u\right)\mathrm{d}u}\,\mathrm{d}v} $$
We can substitute $ \left(u,v\right) $ to $ \left(v,u\right) $ in the right side to get : $$ \int_{a}^{b}{f\left(u\right)\int_{a}^{u}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u}=\int_{a}^{b}{f\left(u\right)\int_{u}^{b}{f\left(v\right)\mathrm{d}v}\,\mathrm{d}u} $$
A: You are on the right track (with the assumption of antiderivative existing, as stated in the comments).
$$\int_a^b f(u)\int_a^u f(v) dv \;du=  \int_a^b f(u)(F(u)-F(a)) du = \int_a^b f(u)F(u) du - F(a)(F(b)-F(a))$$
$$\int_a^b f(u)\int_u^b f(v) dv \;du = \int_a^b f(u)(F(b)-F(u)) du = -\int_a^b f(u)F(u) du +F(b)(F(b)-F(a))$$
All that's left to show is:
$$ \int_a^b f(u)F(u) du = 1/2\left( F(b)^2 - F(a)^2\right) $$ 
Remember that $f=F^\prime$.
A: The left hand integral is of the form
$$
I=\int_{u=a}^{u=b}\int_{v=a}^{v=u}g(u,v)dvdu
$$ 
It has a region of integration given by $a\leq u \leq b$ and $a \leq v \leq u$. This region of integration can be equivalently written as $a \leq v \leq b$ and $v \leq u \leq b$ so a direct application of Fubini's theorem gives
$$
I=\int_{v=a}^{v=b}\int_{u=v}^{u=b}g(u,v)dudv
$$
If the integrand satisfies $g(u,v)=g(v,u)$ for all $u,v$ in the domain of integration then we can swap variables by $u\to v$ and $v\to u$ to obtain
$$
I=\int_{u=a}^{u=b}\int_{v=u}^{v=b}g(v,u)dvdu=\int_{u=a}^{u=b}\int_{v=u}^{v=b}g(u,v)dvdu
$$
The given integrand $g(u,v)=f(u)f(v)$ is one example of such an invariant integrand. Some more would be $g(u,v)=u^2+v^2$ and $g(u,v)=\sin uv+e^{u+v}+u^4+v^4$, and so on. So, the result is more general than for integrands of the form $g(u,v)=f(u)f(v)$.
