KEAM 2012 Math Paper

Given lines are y=x+2,‌‌y=2−x
The intersection point is
2−x=x+2⇒2x=0
⇒‌‌x=0⇒y=2
∴ Intersection point is (0,2) therefore Required shaded region = Area of shaded region AOB+ Area of shaded region BOC
=

0

∫

−2

ydx+

2

∫

0

ydx
=

0

∫

−2

(x+2)dx+

2

∫

0

(2−x)dx
=[‌

x2

2

+2x]−20+[2x−‌

x2

2

]02
=2+2=4sq units
Alternate method
∴ Area of bounded region = Area of △ABC
=‌

1

2

×AC×OB
=‌

1

2

×4×2=4sq units