$\int_0^1\int_x^1f(t) \,dt \,dx=\int_0^1t f(t) \,dt$

Question

Lets suppose that f is continuous in $[0,1]$.
I want to prove that $$\int_0^1\int_x^1f(t) \,dt \,dx=\int_0^1t f(t) \,dt$$ for the left part I set the following function: $$F(y)=\int_0^y f(t)\,dt$$ so from the left part I have that
$$\int_0^1\int_x^1f(t)\,dt\,dx$$ $$=\int_0^1(F(1)-F(x))\,dx$$ $$=F(1)-\int_0^1F(x)\,dx$$ $$=F(1)-F(0)-\int_0^1F(x)\,dx$$
Then, I tried to solve the right part by integrating by parts but I got confussed because of the limits of integration.
$u=t$, $du=dt$, $v=\int f(t)\,dt$, $dv=f(t)$.

Here is were I´am stuck, can you help me?

Answer 1

Change the order of integral.

$ \displaystyle \int_0^1\int_x^1 f(t) \ dt \ dx = \int_0^1\int_0^t f(t) \ dx \ dt$