The CDF is defined as follows:
$$ F(x) = \begin{cases} 0,\qquad x \lt 0 \\[3ex] \frac{x^2}{18}+\frac{x}{6}, \quad 0 \le x \lt 3\\[2ex] 1,\qquad x \ge 3 \end{cases} $$
And i have to calculate the Expectation.
I tried to calculate it in following way:
$\int_0^3{f(x)(\frac{x^2}{18}+\frac{x}{6}})$
where $f(x)$ is the PDF = $\frac{x}{9}+\frac{1}{6}$ which is derivate of CDF.
Now as a result, when i calculate the integral i get $\frac{1}{2}$ which is incorrect. What i´m doing wrong / missing here ?
If $X\ge 0$ as in your case, then the expectation can be directly calculated from the CDF by the formula $$E[X]=\int_{0}^{\infty}(1-F(x))dx$$ which in that case yields $$E[X]=\int_{0}^{3}1-\left(\frac{x^2}{18}+\frac{x}{6}\right)dx+\int_{3}^{+\infty}(1-1)dx$$