find the Integration of the function f(x)=||x-2|+1| in interval [-1,1]? and also find area of Integral function?

if -1=<x<=1 then (x-2)<0 thus mod(x-2)=(2-x)
so mod(mod(x-2)+1)=mod(3-x)
if -1=<x<=1 then (3-x)>0 thus mod(3-x)=3-x
integration=(3x-x²/2)_-1¹
=(3-1/2)-(-3-1/2)
=6
area=integration=6

plot of above function: the function f(x) = x-2+1=x-1 for x>2 = -x+2+1=-x+3 for x<2 } )f(x)dx=)x-1 (between 2 to x(x>2))=(1/2)x^2-x =)-x+3(between x to 2(x<2))=(1/2)(-x^2)+3x