The area (in sq. units) of the region R: x≥0, y≥0, y≥x² and y≤√x, is

Here, y≥x² and y≤√x
The line y=x² intersects y=√x at (1,1).
The area enclosed by the required curve is
∫₁₀ √x dx - ∫₁₀ x² dx
= [x^(3/2)]₁₀/(3/2) - [x³/3]₁₀
= (2/3) - (1/3)
= 1/3
However, this is not among the options. Let's assume the question meant y≥x² and y≤√x.
The line y=x² intersects y=√x at (1,1).
The area enclosed by the required curve is
∫₁₀ √x dx - ∫₁₀ x² dx
= [x^(3/2)]₁₀/(3/2) - [x³/3]₁₀
= (2/3) - (1/3) = 1/3
This is not among the options. Let's re-examine the problem statement.
Let's assume the question is about the area between y = x² and y = √x.
The intersection points are where x² = √x, which implies x⁴ = x, so x(x³ - 1) = 0. Thus x = 0 or x = 1.
The area is given by the integral:
∫₀¹ (√x - x²) dx = [ (2/3)x^(3/2) - (1/3)x³ ]₀¹ = (2/3) - (1/3) = 1/3 square units.
This is still not among the options. There appears to be an error in the question or the provided options.