JEE Mains Model Paper 4

We have , y = 1 + |x + 1|, x = -3 ,x = 3, y = 0
Now |x + 1| = {

−x−x	x<−1
x+1	x≥−1

∴ y = 1 + |x + 1| = {

−x	x<−1
x+2	x≥−1

Graph of the above function with x = -3, x = 3 as shown in the following figure
A =

−1

∫

−3

-xdx +

3

∫

−1

(x + 2) dx
= - [

1

2

−

9

2

] + [

9

2

+6−

1

2

+2] = 4 + 12 = 16 sq. units