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Let f(x)= cases-2, & -2 x 0 \ x-2, & 0<x 2 cases and h(x)=f(|x|)+|f(x)|. Then -22 h(x) dx is equal to:
(1) 2
|x| [0, 2] f(|x|)=|x|-2 (at x=0 also -2)
-2 x 0: h(x)=(-x-2)+|-2|=-x
0<x 2: h(x)=(x-2)+(2-x)=0
[figure]
-22h(x) dx= -20(-x) dx+0= [- x22 ]-20=2 (area of the shaded triangle)
Answer: (1)
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