To find the area bounded by the curve \(y = x(2 - x)\), the x-axis, \(x = 0\), and \(x = 2\), we can set up an integral to calculate the definite integral of the curve over the given interval. The formula for finding the area under a curve is:

\[A = \int_{a}^{b} f(x) \, dx\]

In this case, \(f(x) = x(2 - x)\), \(a = 0\), and \(b = 2\). So, the area \(A\) is given by:

\[A = \int_{0}^{2} x(2 - x) \, dx\]

Let's solve this integral step by step:

\[A = \int_{0}^{2} (2x - x^2) \, dx\]

\[A = \left. \left(x^2 - \frac{x^3}{3}\right) \right|_{0}^{2}\]

Evaluate the expression at the upper and lower limits:

\[A = \left(2^2 - \frac{2^3}{3}\right) - \left(0^2 - \frac{0^3}{3}\right)\]

\[A = \left(4 - \frac{8}{3}\right) - 0\]

\[A = \frac{12 - 8}{3}\]

\[A = \frac{4}{3} \, \text{square units}\]

So, the area bounded by the curve \(y = x(2 - x)\), the x-axis, \(x = 0\), and \(x = 2\) is \(\frac{4}{3}\) square units.