MHB Expected value of a continuous random variable

[rayne1]

Given the PDF:

f(x) = 1/12 , 0 < x <= 3
x/18, 3 < x <= 6
0, otherwise

find the expected value, E(x).

I know how to find the expected value if there was only one interval, but don't how to do it for two.

 

[chisigma]

Given the PDF:

f(x) = 1/12 , 0 < x <= 3
x/18, 3 < x <= 6
0, otherwise

find the expected value, E(x).

I know how to find the expected value if there was only one interval, but don't how to do it for two.

The integral can be devided in two integrals as follows...

$\displaystyle E \{X \} = \frac{1}{12}\ \int_{0}^{3} x\ d x + \frac{1}{18}\ \int_{3}^{6} x^{2}\ dx\ (1)$

Kind regards

$\chi$ $\sigma$

 

[rayne1]

The integral can be devided in two integrals as follows...

$\displaystyle E \{X \} = \frac{1}{12}\ \int_{0}^{3} x\ d x + \frac{1}{18}\ \int_{3}^{6} x^{2}\ dx\ (1)$

Kind regards

$\chi$ $\sigma$

Oh I did try that, so then I must have made a calculation error.