Area of the region bounded by $f(x)$ and $g(x)$? ($f(x)=|2-x|$ and $g(x)=2-|2-x|$) $f(x)=|2-x|$
$g(x)=2-|2-x|$
Find the area of the region bounded by $f(x)$ and $g(x)$.
I was told that you actually need two definite integrals to solve this problem, since it involved absolute value, and $$\int_1^3 [g(x)-f(x)] dx$$ wasn't going to cut it. But I'm not exactly sure why. Instead, I would have to do: $$\int_1^2 [g(x)-f(x)] dx + \int_2^3 [g(x)-f(x)] dx$$ ...or (since a cursory graph sketch reveals symmetry about the line $x=2$), $$2\int_1^2 [g(x)-f(x)] dx$$.
How do you go about solving these absolute value area-between-curves problems? They sound really tricky due to the presence of potential signage errors, and I would like to know how to solve them.
 A: 
You can solve such questions easily by drawing the graph.
After drawing the graph,you can find out the area of the enclosed region, which in this case is a square whose side length is $\sqrt{2}$ units,therefore the area is 2 sq. units.
A: To add onto RamanujanXV's answer I did the following.
I split the area into two parts, where I integrate from 1 to 2, and then again from 2 to 3 because of the equations of the lines. Let the black line be $y=x$ and the red line be $-x+2$. Then we will integrate $\int \text{top-bottom}$. The pink is the area we are currently integrating.
$$\int_1^2 x-(-x+2)dx
\\ =\int_1^22x-2dx = x^2-2x\big|_1^2 = (4-4)-(1-2) = 0--1 = 1
$$

Now we will evaluate the other side using another two linear equations to model the original problem. Let the blue line be $y=-x=4$ and the green line be $y=x-2$. Then we can integrate the second part. The pink is the area we are currently integrating.
$$\int_2^3 -x+4-(x-2)dx = \int_2^3 2x+6dx = -x^2 +6x \big|_2^3
\\ = (-9+18)-(-4+12) = 1$$

Adding these two areas together: $\boxed{\text{area is }1+1=2.}$
To note: The big idea here is to model linear equations that will help you integrate the area you wish. We had two linear equations until x=2, and then we had to change to different linear equations to get the correct area using integration. You could of used the idea of symmetry and computed $2\cdot \int_1^2 x-(-x+2)dx$ and arrived at the same answer.
