If X is uniformly distributed over $(-1, 1)$, find $P(|X| \gt \frac12)$ If X is uniformly distributed over $(-1, 1)$, find $P(|X| \gt \frac12 )$  
I know what the pdf looks like, it is $0$ when $x \gt 1$ and $x \lt -1$, and when $-1 \lt x \lt 1$ it is a constant $\frac12$.
if the question was asking for $P(X \gt \frac12)$ I would do $\int_\frac12^1 \frac12 dx$ and get the answer $\frac14$, but because there is an absolute value around the X in the question I don't know what to do?
Do I have to calculate two integrals and add them?
$\int_\frac12^1 \frac12 dx$ and $\int_{-1}^{-\frac12} \frac12 dx$?
 A: Yes.   You must integrate over the two disjoined intervals of the support where the absolute value of the random variable are greater than an half.   Linearity of Integration means you may sum the integrals over each.
$$\begin{align}
\mathsf P(\lvert X\rvert > \tfrac 1 2)
 & = \displaystyle\int_{\{x\in(-1;1):\lvert x\rvert>1/2\}} f_X(x)\operatorname d x
\\[1ex] & = \int_{(-1;-1/2)\cup(1/2;1)}\tfrac 1 2\operatorname d x
\\[1ex] &  = \int_{-1}^{-1/2} \tfrac 1 2\operatorname d x+ \int_{1/2}^1 \tfrac 12\operatorname d x
\end{align}$$
Which, as hardmath commented, doesn't actually need any calculus to find.   This is because the constant density means we just need to compare interval lengths.   [Important!: the distribution must be uniform for this to work.]
$$\begin{align}
\mathsf P(\lvert X\rvert > \tfrac 1 2)
& = \frac{\lvert(-1;-1/2)\cap(1/2;1)\rvert}{\lvert(-1;-1/2)\cap(1/2;1)\rvert}{\lvert(-1;1)\rvert}
\\[1ex] & = \tfrac 1 2
\end{align}$$
But there's nothing wrong with ensuring you've mastered the concept before moving onto non-uniform distributions.   Good work.
