# Thread: See if you can integrate this reliability function

1. ## See if you can integrate this reliability function

The integral is from t to infinity and R(t)= 2 divided by (2 + t cubed) where t is greater than zero. Obvious methods substitution, integration by parts do not work. i was unable to factorise the denominator into linear factors and therefore it may not be possible to use partial fractions. Your help would be much appreciated. thanks

2. $\displaystyle \int_{t}^{\infty}\frac{2}{2+t^{3}}dt$

You appear to have a dependent limit. Are you sure that is correct?.

3. Originally Posted by mbbx5va2
The integral is from t to infinity and R(t)= 2 divided by (2 + t cubed) where t is greater than zero. Obvious methods substitution, integration by parts do not work. i was unable to factorise the denominator into linear factors and therefore it may not be possible to use partial fractions. Your help would be much appreciated. thanks
are you trying 2 find rate and time ?

4. Originally Posted by mbbx5va2
The integral is from t to infinity and R(t)= 2 divided by (2 + t cubed) where t is greater than zero. Obvious methods substitution, integration by parts do not work. i was unable to factorise the denominator into linear factors and therefore it may not be possible to use partial fractions. Your help would be much appreciated. thanks
$\displaystyle \int_{t}^{\infty}\frac{2}{2+x^{3}} \, dx$

To get the integral, you require a partial fraction decomposition:

$\displaystyle \frac{2}{2+x^{3}} = \frac{2}{(x + a)(x^2 - ax + a^2)} = \frac{A}{x+a} + \frac{B}{x^2 - ax + a^2}$

where $\displaystyle a = 2^{1/3}$ and you get the values of A and B in the usual way.

While this is a pain due to the awkward value of a, it's a straightforward pain.

Then you have to integrate. I get ..... ... a mess that I can't even be bothered typing.

6. I am trying to find the mean life from new of the reliability function R(t)=2/(2+t^3)

7. Originally Posted by mbbx5va2
I am trying to find the mean life from new of the reliability function R(t)=2/(2+t^3)
$\displaystyle R(t) = 1 - \int_0^t f(s) \, ds$ where f(x) is the pdf of X.

Hence $\displaystyle R'(t) = -f(t) \Rightarrow f(t) = - R'(t)$.

Therefore $\displaystyle f(t) = \frac{6t^2}{(2 + t^3)^2}$.

Therefore $\displaystyle \bar{X} = \int_0^{\infty} x f(x) \, dx =$ tough nut. The approximate value is 1.52.