Find the value of the expectation $E(X)$ of the following distribution. 
Let $X$ be a random variable having the distribution function 
  $$F(x)=\begin{cases}
0 &\text{if } x<0\\
\frac{1}{4} & \text{if } 0\leq x<1\\
\frac{1}{3} & \text{if } 1\leq x<2\\
\frac{1}{2} & \text{if } 2\leq x<11/3\\
1& \text{if } x\geq 11/3\end{cases}$$
Find the value of $E(X)$.

Now we know $E(X)=\int_{-\infty}^{\infty}xf(x)dx=\int_{0}^{1}\frac{1}{4}xdx+\int_{1}^{2}\frac{1}{3}xdx+\int_{2}^{11/3}\frac{1}{2}xdx$,
now I am stuck as i don't know what to do with $x\geq 11/3$? Please help me out. Thank you.

This thing isn't a continuous random variable, so the formula you're using doesn't work.
I recommend drawing this thing out. 
The expectation is
$$E[X] = (1-0)\left(1-\frac{1}{4}\right)+(2-1)\left(1-\frac{1}{3}\right) +\left(\frac{11}{3}-2\right)\left(1-\frac{1}{2}\right)=2.25$$

Alternatively, the distribution of $X$ is
\begin{array}{r|ccccc}
x & 0 & 1 & 2 & \frac{11}{3}\\\hline
P(X = x)& \frac{1}{4} & \frac{1}{12} & \frac{1}{6} & \frac{1}{2}
\end{array}
Then,
$$E[X] = 0(1/4)+1(1/12)+2(1/6)+(11/3)(1/2) = 2.25$$
