Draw me a complex math problem

Set x² = √(x).
Square both sides: (x²)² = (√(x))²
x⁴ = x
x⁴ - x = 0
x(x³ - 1) = 0
This gives x = 0 or x³ - 1 = 0, which means x³ = 1, so x = 1.
The points of intersection are at x = 0 and x = 1.

Let's test a value between 0 and 1, for example, x = 0.5.
For y = x², y = (0.5)² = 0.25.
For y = √(x), y = √(0.5) ≈ 0.707.
So, √(x) is greater than x² in the interval [0, 1].

The area A is given by the integral of the upper curve minus the lower curve from the lower bound to the upper bound of intersection.
A = ∫ from 0 to 1 of (√(x) - x²) dx

A = ∫ from 0 to 1 of (x^(1/2) - x²) dx
Integrate term by term:
A = [ (x^(3/2))/(3/2) - (x³)/3 ] from 0 to 1
A = [ (2/3)x^(3/2) - (1/3)x³ ] from 0 to 1

A = [ (2/3)(1)^(3/2) - (1/3)(1)³ ] - [ (2/3)(0)^(3/2) - (1/3)(0)³ ]
A = [ (2/3) - (1/3) ] - [ 0 - 0 ]
A = (1/3)

The area of the region bounded by the curves y = x² and y = √(x) is (1/3) square units.