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

I will assume that $f$ is absolutely integrable so that we can apply Fubini. Using Iverson Brackets:
$$
\begin{align}
\int_0^1\int_x^1f(t)\,\mathrm{d}t\,\mathrm{d}x
&=\int_0^1\int_0^1f(t)[t\gt x]\,\mathrm{d}t\,\mathrm{d}x\tag1\\
&=\int_0^1\int_0^1f(t)[t\gt x]\,\mathrm{d}x\,\mathrm{d}t\tag2\\
&=\int_0^1tf(t)\,\mathrm{d}t\tag3
\end{align}
$$
Explanation:
$(1)$: write the inner integral using the indicator function of $t\gt x$
$(2)$: change order of integration
$(3)$: evaluate the integral in $x$
