KEAM 2010 Math Paper

Given curves x=y2−2 and y=xThus, interection point are (−1,1) and ‌‌(2,−2)
We are to find the area of shaded part
Area of ABC=

−1

∫

−2

√x+2dx
=[‌

2

3

(x+2)3/2]−2−1=‌

2

3

sq unit
Area of BCO=

0

∫

−1

−xdx=(−‌

x2

2

)−10
=‌

1

2

sq unit
Area of ADO
=‌

4

3

√2
Area of ODE= area of ODEF− area of OFE

2

∫

0

√x+2dx−

2

∫

0

(−x)dx
={‌

2

3

(x+2)3∕2}02−(−‌

x2

2

)02
=(‌

16

3

−‌

4√2

3

)−(2)
(neglecting the negative sign)
=‌

16

3

−‌

4√2

3

−2sq unit
∴ Required area
=‌

2

3

+‌

1

2

+‌

16

3

−2
=‌

27

6

=‌

9

2

sq unit