probability with an absolute value given a probability density function

Question

Having the p.d.f.

$f(x)=\frac{1}{2}e^{-|x|}\mathbb{I}_{(-\infty,\infty)}(x)$

Determine the probability of $P(1\leq |x|\leq 2 )$

and what I did was

\begin{equation} \begin{split} P(1\leq |x| \leq2)&=1-P(-1\leq x \leq 1)+P(-2\leq x \leq 2)\\ &=1-\frac{1}{2}\int_{-1}^{1}e^{-|x|}dx+\frac{1}{2}\int_{-2}^{2}e^{-|x|}dx\\ \end{split} \end{equation} which by doing the maths I get that \begin{equation} \begin{split} P(1\leq |x| \leq2)&=1-\frac{1}{2}\Big[e^{0}-e^{-1}\Big]-\frac{1}{2}\Big[-e^{-1}+e^{0}\Big]+\frac{1}{2}\Big[e^{0}-e^{-2}\Big]+\frac{1}{2}\Big[-e^{-2}+e^{0}\Big]\\ &=1+e^{-1}-e^{-2}=1.2325 \end{split} \end{equation} But my problem here is that I don't have an idea where do I get that extra 1 in my result.

Answer 1

$P(1\leq |x| \leq 2)=P(|x| \leq 2)-P(|x| < 1)$ so the very first line in your calculation is wrong.

Answer 2

$$2 \int\limits_{x=1}^2 e^{-x}\ dx = \frac{e-1}{e^2}$$