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Math Help - See if you can integrate this reliability function

  1. #1
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    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
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  2. #2
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    \int_{t}^{\infty}\frac{2}{2+t^{3}}dt

    You appear to have a dependent limit. Are you sure that is correct?.
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  3. #3
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    Quote Originally Posted by mbbx5va2 View Post
    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 ?
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  4. #4
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    Quote Originally Posted by mbbx5va2 View Post
    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
    \int_{t}^{\infty}\frac{2}{2+x^{3}} \, dx

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

    \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 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.
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    Hi, thanks for your reply it has been very helpful.
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  6. #6
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    I am trying to find the mean life from new of the reliability function R(t)=2/(2+t^3)
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  7. #7
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    Quote Originally Posted by mbbx5va2 View Post
    I am trying to find the mean life from new of the reliability function R(t)=2/(2+t^3)
    R(t) = 1 - \int_0^t f(s) \, ds where f(x) is the pdf of X.

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

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

    Therefore \bar{X} = \int_0^{\infty} x f(x) \, dx = tough nut. The approximate value is 1.52.
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  8. #8
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    Thanks for your reply it has been very helpful.
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