The area of the region given by(x, y): y ≥ x^2, y ≤ |x| + 2 _________…

The correct answer is Option (3) → $\frac{20}{3}$ sq.units

$y ≥ x^2, y ≤ |x|+2$

I = II

By symmetry

in region II → $y=x+2$

$y=x^2$

so $x^2=x+2$

$x=2,-1$

$x≠-1$

so $x=2$ only

$y=4$

so area = 2 × ar(II)

$=2×(\int\limits_0^4\sqrt{y}dy-\int\limits_2^4y-2dy)$

$=2\left(\left[\frac{2}{3}y^{\frac{3}{2}}\right]_0^4-\left[\frac{y^2}{2}-2y\right]_2^4\right)$

$=\frac{20}{3}$ sq. units