The problem provides the equations for the two curves, y1 = sqrt(ax) and y2 = (x²)/(a), and the integral setup:

The problem provides the equations for the two curves, $y_1 = \sqrt{ax}$ and $y_2 = \frac{x^2}{a}$, and the integral setup:

$$A = \int_{0}^{a} (y_1 - y_2) \,dx$$

Substitute the expressions for $y_1$ and $y_2$:

$$A = \int_{0}^{a} \left(\sqrt{ax} - \frac{x^2}{a}\right) \,dx$$

$$\sqrt{ax} = \sqrt{a} \cdot \sqrt{x} = \sqrt{a} x^{1/2}$$ $$\frac{x^2}{a} = \frac{1}{a} x^2$$

So the integral becomes:

$$A = \int_{0}^{a} \left(\sqrt{a} x^{1/2} - \frac{1}{a} x^2\right) \,dx$$

Recall the power rule for integration: $\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$.

$$\int \sqrt{a} x^{1/2} \,dx = \sqrt{a} \frac{x^{1/2+1}}{1/2+1} = \sqrt{a} \frac{x^{3/2}}{3/2} = \frac{2}{3} \sqrt{a} x^{3/2}$$ $$\int \frac{1}{a} x^2 \,dx = \frac{1}{a} \frac{x^{2+1}}{2+1} = \frac{1}{a} \frac{x^3}{3} = \frac{x^3}{3a}$$

$$A = \left[ \frac{2}{3} \sqrt{a} x^{3/2} - \frac{x^3}{3a} \right]_{0}^{a}$$

Substitute the upper limit $x=a$:

$$\left( \frac{2}{3} \sqrt{a} (a)^{3/2} - \frac{(a)^3}{3a} \right)$$

Simplify the terms:

$$\frac{2}{3} a^{1/2} a^{3/2} - \frac{a^3}{3a} = \frac{2}{3} a^{(1/2 + 3/2)} - \frac{a^2}{3} = \frac{2}{3} a^{4/2} - \frac{a^2}{3} = \frac{2}{3} a^2 - \frac{a^2}{3}$$

Substitute the lower limit $x=0$:

$$\left( \frac{2}{3} \sqrt{a} (0)^{3/2} - \frac{(0)^3}{3a} \right) = 0 - 0 = 0$$

Subtract the lower limit value from the upper limit value:

$$A = \left( \frac{2}{3} a^2 - \frac{a^2}{3} \right) - 0$$ $$A = \frac{2a^2 - a^2}{3} = \frac{a^2}{3}$$

The final answer is $\frac{a^2}{3}$.