For a continuous random variable, the probability over an interval is calculated by taking the definite integral of the probability density function (PDF) over that interval.

1. Set up the integral to find \(P(X>2)\):
The PDF is given as \(f(x) = \frac{x}{18}\) on the interval \(0 \le x \le 6\). To find \(P(X>2)\), we integrate \(f(x)\) from the lower limit \(x=2\) to the function's upper limit, \(x=6\).
\[P(X>2) = \int_{2}^{6} f(x) \,dx = \int_{2}^{6} \frac{x}{18} \,dx\]
2. Evaluate the definite integral:
First, find the antiderivative of the function \(\frac{x}{18}\):
\[\int \frac{x}{18} \,dx = \frac{1}{18} \int x \,dx = \frac{1}{18} \left( \frac{x^2}{2} \right) = \frac{x^2}{36}\]
Next, apply the limits of integration (from 2 to 6) using the Fundamental Theorem of Calculus:
\[\begin{aligned} P(X>2) &= \left[ \frac{x^2}{36} \right]_{2}^{6} \\ &= \left( \frac{6^2}{36} \right) - \left( \frac{2^2}{36} \right) \\ &= \frac{36}{36} - \frac{4}{36} \\ &= \frac{36 - 4}{36} \\ &= \frac{32}{36} \end{aligned}\]
3. Simplify the result:
Reduce the fraction to its simplest form by dividing the numerator and denominator by their greatest common divisor, which is 4.
\[\frac{32}{36} = \frac{8}{9}\]
Result:
The value of \(P(X>2)\) is \(\frac{8}{9}\).