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Math Help - trouble with e

  1. #1
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    trouble with e

    Hi

    f(x) = e^-x/ (1-e^-2) if 0<= x <= 2

    so the question is to find the mathematical expectation of X - ie calculate

    integral sign ("S") b=2; a=0 x f(x) dx

    Now I am not completely sure where the x came from so a clue there would be helpful.

    But really I am not sure what to do with this - even if the x shouldn't be there I am not sure if I should use partial integration since I can't tell what should be
    u and what should be dv when both have e's to the power so I don't know how that would make them dv able.

    All help would be appreciated.

    Beginner Calculus
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Do you know what the expected value is? Expected value is the weighed average of the possible values a random variable can assume. Here f(x) is the probability density function (i.e. the weight). The analogue for a discrete probability distribution would work like this : suppose you had a random variable which took the values 1, 2, 3 with respective probabilities 0.3, 0.3, 0.4. Then the expected value would be

    \sum x P(x) = 1 \times 0.3 + 2 \times 0.3 + 3 \times 0.4 = 2.1

    For a continuous probability distribution the sum is replaced by an integral

    \int_D x f(x) \ dx

    where D is the domain of the random variable. You can think of the expected value of a real random variable as the center of mass of a rod whose mass distribution is the probability density function.

    To answer your other question : yes, that integral would be done by parts. Set u=x and dw=e^{-x} dx.
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  3. #3
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    still lost

    Okay so if u = x then du = dx

    if dv = e^-1 then v = e^-1

    so I have S u (v^u/ 1-v^2) and I am lost

    Please explain where I go from here.

    Thanks

    Calculus Beginner
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Why do you set dv=e^{-1}? That is not even a function of x.

    You want to set dv=e^{-x} (so v=-e^{-x}) or dv=-e^{-x} (so v=e^{-x}). In my calculations below I use the first approach.

    I'll do the indefinite integral and I'll let you figure out the details.

    \int xe^{-x}\ dx = \int x\ d(-e^{-x}) = x(-e^{-x}) - \int d(x)(-e^{-x})

    = -xe^{-x} - \int (-e^{-x})\ dx = -xe^{-x}-e^{-x}

    Now we can verify the answer : \frac{d}{dx}(-xe^{-x}-e^{-x}) = -e^{-x}+xe^{-x}+e^{-x} = xe^{-x}

    so it's good.
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