The probability density function (PDF) of a continuous random variable $X$ is a function $f_X(x)$ that gives the probability that $X$ takes on a value between $x$ and $x + dx$, for an infinitesimally small value of $dx$. The total probability that $X$ takes on any value is 1, so the PDF must satisfy the condition $\int_\mathbb{R} f_X(x) dx = 1$.
In this case, the PDF is given by
$$f_X(x) = \begin{cases} K(5x – 2x^2) & \text{if } 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$$
The value of $K$ is not specified, but it can be found by integrating the PDF over the entire real line. This gives
$$1 = \int_\mathbb{R} f_X(x) dx = K \int_0^2 (5x – 2x^2) dx = K \left[ \frac{5x^2}{2} – \frac{2x^3}{3} \right]_0^2 = 4K$$
Therefore, $K = \frac{1}{4}$.
The probability that $X > 1$ is the area under the PDF to the right of $x = 1$. This area can be found by integrating the PDF from $1$ to $2$. This gives
$$P(X > 1) = \int_1^2 f_X(x) dx = \int_1^2 \frac{1}{4} (5x – 2x^2) dx = \frac{5x^2}{8} – \frac{x^3}{12} \bigg|_1^2 = \frac{1}{8} – \frac{1}{12} = \boxed{\frac{3}{14}}$$