Finding $cdf$ of the sample minimum, $X_{(1)}$ 
Consider iid random variables $X_1$ and $X_2$, having $pdf$ $$f_X(x) =
 4(1−2x)I_{(0,1/2)}(x)$$ Give the $cdf$ of the sample minimum,
  $X_{(1)}$.

$$\begin{align*}
F_{X(1)}(x) 
&= P(X_{(1)} \leq x) \\\\
&= 1 - P(min{\{X_1, X_2}\} \gt x) \\\\
&= 1 - P(X_1 \gt x, X_2 \gt x) \\\\
&= 1 - P(X_1 \gt x)\cdot P(X_2 \gt x) \\\\
&= 1 - [1-F_X(x)]^2 \\\\
&= 1 - [1-\int4(1-2x)]^2 \\\\
&= 1 - [1-(4x-4x^2)]^2 \\\\
\end{align*}$$
Did I do this correctly?

seems good to me:
Just a baseline check: $F(1/2) = 1$, and as long as you understand that this is valid for $x \in (0, 1/2)$ we are ok. What would the values of $F$ be for negative $x$ and $x \ge 1/2$?
