Calculate the $P\left(X_{(1)}+X_{(2)}\leq1\right)$ I got the following question:

Let $X_{(1)},X_{(2)},X_{(3)}$ be statistic order of three independent random variables $X_1,X_2,X_3$ with uniform distribution in $(0,1)$. Find the $P\left(X_{(1)}+X_{(2)}\leq1\right)$.

In the solution they got:

I'm trying to figure why it's true. I'm familiar with the trick:
$$
f_{X_{(1)}X_{(2)}}(x,y)\cdot dxdy=P\left(x\leq X_{(1)}\leq x+dx,y\leq X_{(2)}\leq y+dy\right)
$$
But how did they got the result?

If $(X_n)$ is a sequence of i.i.d. $U(0,1)$ random variables, then for $i<j$ the joint density of $(X_{(i)},X_{(j)})$ is
$$
f_{(i),(j)}(s,t) = n!\frac{s^{i-1}}{(i-1)!}\frac{(t-s)^{j-i-1}}{(j-i-1)!}\frac{(1-t)^{n-j}}{(n-j)!}\cdot\mathsf 1_{\{0<s<t<1\}}.
$$
So we have
$$
f_{(1),(2)}(s,t) = 6(1-t)\cdot\mathsf 1_{\{0<s<t<1\}}.
$$
It follows that the density of $X_{(1)}+X_{(2)}$ is
\begin{align}
f_{(1)+(2)}(t) &= \int_{\mathbb R}f_{(1),(2)}(s,t-s)\ \mathsf ds\\
&= \left(\int_0^{\frac t2}6(1-(t-s))\ \mathsf ds\right)\mathsf 1_{(0,1)}(t)+ \left(\int_{t-1}^{\frac t2}6(1-(t-s))\ \mathsf ds\right)\mathsf 1_{(1,2)}(t)\\
&= 3t\left(1-\frac34t\right)\mathsf 1_{(0,1)}(t) + \frac{3}{4} (2-t)^2\mathsf 1_{(1,2)}(t).
\end{align}
We compute the desired probability by integrating this density over $(0,1)$:
\begin{align}
\mathbb P(X_{(1)}+X_{(2)}<1) &= \int_{(0,1)}f_{(1)+(2)}(t)\ \mathsf dt\\
&= \int_0^13t\left(1-\frac34t\right)\ \mathsf dt\\
&=\frac34.
\end{align}
