$\int_0^1\int_x^1f(t) \,dt \,dx=\int_0^1t f(t) \,dt$ 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?
 A: Edit : since OP doesn't precise what is the regularity of $f$, I went ahead and assumed it is continuous.
Let $F(x) = \int_x^1 f(t)\,\text dt$. Then, $F$ is $C^1$ and $F'(x) = -f(x)$. Integration by part gives :
$$\int_0^1 F(x) \,\text dx = [xF(x)]_0^1 - \int_0^1xF'(x) \, \text dx$$
Using $F(1) = 0$, we get :
$$\int_0^1\int_x^1f(t)\, \text dt=\int_0^1 x f(x)\,\text dx$$
A: 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$
A: 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 $x$
A: \begin{align}
& \int_0^1 \left( \int_x^1 \cdots\,dt \right) \,dx \\[8pt]
= {} & \iint\limits_{t,x\,:\,0\,<\,x\,<\,t\,<\,1} \cdots \,d(t,x) \\[8pt]
= {} & \int_0^1 \left( \int_0^t \cdots \, dx \right) \,dt
\end{align}
