Question: The question is in the description. 
One of the answers I am getting is $\int\_0^1 {x (1-x) }=1/6$. 

The other answer I am getting is there are only two possibilities: $x\_1>x\_2$ and $x\_1<x\_2$. Hence probability is 1/2

Any help would be great.

Answer: 
If $X\_1, X\_2 \sim U(0,1)$ are uniformly distributed and independent, then the joint density is $f\_{X\_1X\_2}(x\_1,x\_2) = 1$ and

$$P(X\_1 < X\_2) = \int\_0^1 \int\_0^{x\_2} \, dx\_1 \, dx\_2 = 1/2.$$
If $X\_1,X\_2 \sim U(0,1)$ are uniformly distributed and independent, we have by the elementary law of total probability

$$
\mathbb{P}(X\_1 < X\_2) = \int\_{-\infty}^{\infty}\mathbb{P}(X\_1 < X\_2 | X\_2 = s)f\_{X\_2}(s)ds = \int\_{0}^{1}\mathbb{P}(X\_1 < s)ds = \int\_{0}^{1}sds = \left[\frac{1}{2}s^2 \right]\_{0}^{1} = \frac{1}{2}
$$

Where we have used the result that for $s \in [0,1]$ we have $f\_{X\_2}(s)$ is $1$ (and $0$ otherwise), and that $\mathbb{P}(X\_1 < s) = s$.