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    (Original post by SubAtomic)
    :emo: had to look at evaluating integrals with infinity, so I get this

    \displaystyle \frac{\lambda \cdot e^{-\infty}}{t-\lambda}- \frac{\lambda}{t- \lambda}}

    \displaystyle = \frac{\lambda \cdot e^{-\infty} - \lambda}{t- \lambda}

    Is this right? If so I think I may just about get it
    Well have you understood why M(t) is always positive now?
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    (Original post by Smaug123)
    Pretty much, yep - the infinite integral should technically be evaluated as a limit, but this works fine if you just use lim(e^-x) = 0 as x->inf.
    Yeah I did all the lim stuff here but didn't want to tex it so that is what I expected, so

    \displaystyle = \frac{\lambda \cdot e^{-\infty} - \lambda}{t- \lambda} \equiv \frac{\lambda}{\lambda -t}}
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    (Original post by shamika)
    Well have you understood why M(t) is always positive now?
    Possibly, because the condition t < lambda means if lamda negative then Mx(t) positive and if lambda positive t < lambda means Mx(t) positive ?

    Is that just waffle? Yes because it is given that lambda > 0 hahaha

    Am I a lost cause
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    (Original post by SubAtomic)
    Possibly, because the condition t < lambda means if lamda negative then Mx(t) positive and if lambda positive t < lambda means Mx(t) positive ?

    Is that just waffle? Yes because it is given that lambda > 0 hahaha
    Pretty much waffle; isn't lambda restricted to be positive?

    OK lets go back to the integral you've evaluated. Why have you got e^{-\infty} rather than e^{\infty}?

    And once you've understood that, tell me what happens to e^{-x} as x becomes very large.

    Finally, putting all of that together you'll get the expression on the page you've scanned.
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    (Original post by shamika)
    Pretty much waffle; isn't lambda restricted to be positive?
    Lol, thought so.

    (Original post by shamika)
    OK lets go back to the integral you've evaluated. Why have you got e^{-\infty} rather than e^{\infty}
    Because t < lambda so F(b) has the e^{- \infty} in the numerator.

    (Original post by shamika)
    And once you've understood that, tell me what happens to e^{-x} as x becomes very large.
    Tends to 0

    (Original post by shamika)
    Finally, putting all of that together you'll get the expression on the page you've scanned.
    Yep
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    Many thanks to you all :yay:
 
 
 
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