The area of the plane region bounded by the curve `x = y^(2)-2` and the line y = - x is (in sq units)

Options

A. `(13)/(3)`
B. `(2)/(5)`
C. `(9)/(2)`
D. `(5)/(2)`

Answer

C

Explanation

Given curvel are `x=y^(2)-2` and `y=-x`
On solving these two curves
`y=-(y^(2)-2)`
`implies y^(2)+y-2=0`
`impliesy^(2)+2y-y-2=0`
`implies y(y+2)-+(y+2)=0`
`rArr(y+2)(y-1)=0`
y = 1, -2
When y = 1, then x = - 1
When y = - 2 then x = 2
Therefore intersection points of both the curves are (-1, 1) are (2, -2)

Now, Required area
`=2underset(-2)overset(-1)intsqrt(x+2)dx+underset(-1)overset(2)int(-x+sqrt(x+2))dx`
`=(4)/(3)[(x+2)^(3//2)]_(-2)^(-1)+[(-x^(2))/(2)+(2)/(3)(x+2)^(3//2)]_(-1)^(2)`
`(4)/(3)(1)^(3//2)+[(-(2)^(2))/(2)+(2)/(3)(4)^(3//2)]-[-((-1)^(2))/(2)+(2)/(3)(1)^(3//2)]`
`(4)/(3)-2+(2)/(3).8+(1)/(2)-(2)/(3)=(4+16-2)/(3)-2+(1)/(2)=6-(3)/(2)=(9)/(2)`