How to solve this using definite integral? I already have asked $2$ questions on similar topics. In that questions I have equations of parabola and line. And equations have $y$ and $x$ variable.
But in this question I have functions.
$$f(x) = |x| - 1 \quad\mbox{and}\quad g(x) = 1 - |x|.$$
a) Sketch their graphs.
b) Using integration find the area of the bounded region. 
In other two questions I know how to find values of $x$ and $y$ using substitution. But in this question because of $f(x)$ and $g(x)$. I am clueless how to start.
Please provide answer in detail.
My attempt is only.
$f(x) = x - 1, -x - 1$
$g(x) = 1 - x, 1 + x$.
 A: Hint for f (x) shift the graph of $|x|$ by 1  towards right. For other function use $g (x)=1+x $ when x is negative and $1-x $ for positive. Then find the intersection with $f (x) $and then the area
A: I presumed you know how to sketch the region. Then
$$A=2\int_{0}^1[(1-x)-(x-1)]dx=\int_0^1(4-4x)dx=2.$$
In set notation, the region $R$ is given by
$$R=\{(x,y):-1\leq x\leq 1,f(x)\leq y\leq g(x)\}.$$
The line $x=0$ or known as the $y$-axis serves the line of symmetry.
Okay,this is the graph of the region $R$.
A: Normally when finding area we do $\int_{a}^{b} f(x)-g(x)\text{d}x$ where f is the "top" function along the interval. So looking at a sketch of the region g(x) is on top over the interval and the interval is from -1 to 1. So our integral is $\int_{-1}^{1}1-\left | x\right |-(\left | x \right |-1)\text{d}x=$ $\int_{-1}^{1} 2-2\left |x\right |\text{d}x$ intuitively we can see that $\int \left | x\right |\text{d}x=\frac{\left |x\right |x}{2}+C$ from the piecewise definition of $|x|$ so using this just makes the integral into a pretty standard definite integral. I personally prefer this way because you do not need to figure out where the sign changes and do multiple integrals. However this only works for absolute value functions in the form $\left |ax+b\right |$ with higher powers and other functions you do need multiple integrals. Hope this helps a bit 
