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    Hi, would anybody be able to tell me the integration topic or method to solve this question described by the mark scheme, as I didn't have a clue how to answer the question and need more practise.
    The question is 8a from the AQA C3 January 2013 paper here:

    And the mark scheme being

    I usually just read the mark scheme and work it out but its using k for some reason as well as the exponential 1 just disappearing, if anyone can tell me the integration type this answer requires it will be greatly appreciated, thank you
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    'Recognition'. You know that the differential of e^(kx) = ke^(kx) so you 'work backwards'.

    Also, how can you write x^(1-a) as x^(...) * x^(...)? Then for e^(1-a) it will be the same logic.

    (Original post by Sayless)
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    (Original post by Sayless)
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    As per SeanFM - you should recognise that \displaystyle \int f'(x) e^{f(x)} \, \mathrm{d}x = e^{f(x)} + c
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    (Original post by SeanFM)
    'Recognition'. You know that the differential of e^(kx) = ke^(kx) so you 'work backwards'.

    Also, how can you write x^(1-a) as x^(...) * x^(...)? Then for e^(1-a) it will be the same logic.
    yeah so i simplify it to:
    integral of e x e^(-2x)
    now what do i do I don't know how to approach this types of questions even after I simplify it to that,
    I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together
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    (Original post by Sayless)
    yeah so i simplify it to:
    integral of e x e^(-2x)
    now what do i do I don't know how to approach this types of questions even after I simplify it to that,
    I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together
    e is just a number so you need only integrate e \int_0^{\ln 2} e^{-2x} \, \mathrm{d}x.
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    (Original post by Sayless)
    yeah so i simplify it to:
    integral of e x e^(-2x)
    now what do i do I don't know how to approach this types of questions even after I simplify it to that,
    I know how to do e^2x or 2e^4x differentiation and integration but I don't know when there are 2e's timesed together
    e is a constant and it does not depend on x.
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    (Original post by Zacken)
    e is just a number so you need only integrate e \int_0^{\ln 2} e^{-2x} \, \mathrm{d}x.
    ah right, I thought that if I took e out of the integrand it would mean I would have to take e out of e^(-2x)
    didn't think it worked that way but now it makes sense if its timsing, when I did it I did
    integral of e/e^(2x)
    which i couldnt make sense of
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    (Original post by Sayless)
    ah right, I thought that if I took e out of the integrand it would mean I would have to take e out of e^(-2x)
    didn't think it worked that way but now it makes sense if its timsing, when I did it I did
    integral of e/e^(2x)
    which i couldnt make sense of
    Have you managed to solve the problem now?
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    (Original post by Zacken)
    Have you managed to solve the problem now?
    I solved it but I had to keep e outside the integrand the whole time and the integral,

    I don't know what is e*-1/2(e)^-2x
 
 
 
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