Let F and G be two real-valued function defined on R (the set of all real numbers)
`F(x)={:{(,|x+1|,),(,x,):}`{:(,xle0,),(,xgt0,):}` G(x) ={:{(,|x|+1,),(,-|x-2,):}`{:(,xle1,),(,xgt1,):}` and `H(x)=F(x)+G(x)`.
The value of definite integral `int_(-2)^(2) H(x) dx` is equal to

Options

A. `4`
B. `7`
C. `8`
D. `12`

Answer

C

Explanation

We have, `F (x)={:{(-,(x+1),),(,(x+1),),(,(x),):}{:(,xlt-1,),(,-1 lex le 0,),(,x gt 0,):}`
and `G(x) ={:{(,-x+1,),(,x+1,),(,x-2,),(,2-x,):}{:(,-oo lt x le0,),(,0ltxle1,),(,1ltxlt2,),(,xge2,):}`
`H(x)=F(x)+G(x)={:{(,-(x-1-x+1=-2x),),(,(x+1)-x+1=2,),(,x+(x+1)=2x+1,),(,x+(x-2)=2x-2,),(,x+2-x=2,):} {:(,-ooltxle-1,),(,-1lexle0,),(,0lexle1,),(,1ltxlt2,),(,xge2,):}`
Graph of H(x)
Clearly from the graph, H(x) is discontinous at
x=0 and x=1, i.e., 2 points of discontinuity
` therefore A & B`
Graphically,
Area under curve from `-2 to 2`
`=((1)/(2)xx2xx1xx1)+(2+1)+((1)/(2)xx2xx1xx1)+((1)/(2)xx2xx1)`
`=(1+2)+2+(1+1)+1`
`=3+2+2+1=8`