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Math Help - Integral help

  1. #1
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    Integral help

    Need to find the integral of e^[t^(2) - t]dt
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by element View Post
    Need to find the integral of e^[t^(2) - t]dt
    This cannot be expressed in terms of simple functions.

    Integrate[E^(x^2 - x), x] ==
    (Sqrt[Pi]*Erfi[-1/2 + x])/(2*E^(1/4))
    http://reference.wolfram.com/mathematica/ref/Erfi.html
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    This cannot be expressed in terms of simple functions.

    Integrate[E^(x^2 - x), x] ==
    (Sqrt[Pi]*Erfi[-1/2 + x])/(2*E^(1/4))
    Erfi - Wolfram Mathematica
    thanks. hmm. i probably did something wrong. here is the problem i am working on. we are supposed to use an integrating factor for these linear first order differential equations.

    dP/dt + 2tP = P + 4t -2

    i manipulated this into "standard form" ( dy/dx + P(x)y = f(x) )

    dP/dt + (2t - 1)P = 4t - 2

    then the integrating factor would be e^integral of [2t -1]dt, and that would be e^[t^(2) - t] correct?

    now my differential equation looks like this

    e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt

    i separated the right side into
    4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt

    now im stuck on integrating that.
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  4. #4
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    Quote Originally Posted by element View Post
    thanks. hmm. i probably did something wrong. here is the problem i am working on. we are supposed to use an integrating factor for these linear first order differential equations.

    dP/dt + 2tP = P + 4t -2

    i manipulated this into "standard form" ( dy/dx + P(x)y = f(x) )

    dP/dt + (2t - 1)P = 4t - 2

    then the integrating factor would be e^integral of [2t -1]dt, and that would be e^[t^(2) - t] correct?
    Yes, that is correct.

    now my differential equation looks like this

    e^[t^(2) - t]y = integral of [ (4t-2)e^[t^(2) - t] ]dt

    i separated the right side into
    4 integral of te^[t^(2) - t]dt -2 integral of e^[t^(2) - t]dt

    now im stuck on integrating that.
    Don't separate! Let u= t^2- t. The integral becomes 2\int e^u du
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that is correct.


    Don't separate! Let u= t^2- t. The integral becomes 2\int e^u du
    thank you very much! i cant believe i didn't see that. thanks.
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