Symmetry of double integral

Question

I am a little embarassed that I cannot figure out this on my own, but oh well...

Let $f \colon [a,b] \to \mathbb{R}$ be continuous. Why is it true that $$ \int_0^1 f(x) \int_0^x f(y) dy \,dx = \int_0^1 f(x) \int_x^1 f(y)dy\,dx~~~~ ? $$

I can see why this should be correct. I believe that I should use the substitution rule, but do not know which substitution to use.

Answer 1

It's "Fubini's theorem" about reducing double integrals, resp., the inverse operation. If $$T:=\{(x,y)\>|\>0\leq y\leq x\leq1\}\ ,$$ and $g: \ (x,y)\mapsto g(x,y)$ is a reasonable function defined on $T$ then $$\int_T g(x,y)\>{\rm d}(x,y)=\left\{\eqalign{&\int_0^1\int_0^x g(x,y)\>dy\>dx \cr &\int_0^1\int_y^1 g(x,y)\>dx\>dy\cr}\right.\quad.$$ In the case at hand $g(x,y)=f(x)\,f(y)$, so that we obtain (taking constant factors out of the inner integrals) $$\int_0^1 f(x)\int_0^x f(y)\>dy\>dx=\int_0^1 f(y)\int_y^1 f(x)\>dx\>dy\ .$$

Answer 2

HINT : let $F$ be a primitive of $f$, you should be able to solve the integrale and see that both sides are equal

Answer 3

Step $1:$ Change the order of the integration in LHS to get $$ \int_0^1\int_y^1 f(x)f(y)dx dy $$ Step $2:$ Switch the variable names. $$ \int_0^1\int_x^1 f(y)f(x)dy dx $$