8.3 Area Between Two Curves

Intersection Points: Substitute $y=x^2$ into $y^2=x$: $(x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0$. Real roots: $x = 0$ and $x = 1$. Points: $(0,0)$ and $(1,1)$.

Identify Upper/Lower: In the interval $(0, 1)$, for a test point $x = 1/4$: $y = x^2 = 1/16 = 0.0625$. $y = \sqrt{x} = 1/2 = 0.5$. So $y = \sqrt{x}$ (from $y^2=x$) is the upper curve.

Integral: $A = \int_{0}^{1} (\sqrt{x} - x^2) dx$ $A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_{0}^{1}$ $A = \left( \frac{2}{3}(1) - \frac{1}{3} \right) - 0 = \frac{1}{3} \text{ sq units}$