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Integration - Where have I made a mistake? Watch

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    By the way, I'm using substitution. The answer is around 33,7, but my answer is no where near...


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    (Original post by ps1265A)
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    By the way, I'm using substitution. The answer is around 33,7, but my answer is no where near...


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    9 and 4 are the limits for u

    You changed back into x so you should have used the x limits

    Or just left it in u and use the u limits
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    (Original post by ps1265A)
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    By the way, I'm using substitution. The answer is around 33,7, but my answer is no where near...


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    You used the wrong limits. After you integrated, you changed it back in terms of x. If you do this, you need to change the limits back.
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    Very Important Poster
    Third to last line - you've swapped u for x but you are still using the limits for u.
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    (Original post by TenOfThem)
    9 and 4 are the limits for u

    You changed back into x so you should have used the x limits

    Or just left it in u and use the u limits
    Thanks, could you help me with this one?

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    (Original post by ps1265A)
    Thanks, could you help me with this one?

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    The actual integration

    If you differentiate 2uln(1+u) wrt u is it correct ?
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    (Original post by TenOfThem)
    The actual integration

    If you differentiate 2uln(1+u) wrt u is it correct ?
    How can I integrate this fraction then?


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    (Original post by TenOfThem)
    The actual integration

    If you differentiate 2uln(1+u) wrt u is it correct ?
    I implemented the same method as...Name:  ImageUploadedByStudent Room1421950291.479975.jpg
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    (Original post by ps1265A)
    I implemented the same method as...Name:  ImageUploadedByStudent Room1421950291.479975.jpg
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Size:  139.3 KB


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    No you didn't

    In that example the numerator was the differential of the denominator


    In \dfrac{2u}{u+1} that is not the case

    You need to split this using long division or partial fraction
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    (Original post by TenOfThem)
    No you didn't

    In that example the numerator was the differential of the denominator


    In \dfrac{2u}{u+1} that is not the case

    You need to split this using long division or partial fraction
    I've got to this stage:
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    Ended up with another awkward fraction to integrate... In this case, I can't use partial fractions


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    (Original post by ps1265A)
    I've got to this stage:
    Name:  ImageUploadedByStudent Room1421951780.862908.jpg
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    Ended up with another awkward fraction to integrate... In this case, I can't use partial fractions


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    2u divided by u is 2 not 1


    What can't you use partial fractions
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    (Original post by TenOfThem)
    2u divided by u is 2 not 1


    What can't you use partial fractions
    2u/u+1
    A/u+1 + B/u+1

    I'm stuck when I use partial fractions and when x=1, both A and B will cancel


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    (Original post by ps1265A)
    2u/u+1
    A/u+1 + B/u+1

    I'm stuck when I use partial fractions and when x=1, both A and B will cancel


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    It would be A + B/(u+1)
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    (Original post by TenOfThem)
    It would be A + B/(u+1)
    So 2u = A + B


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    (Original post by ps1265A)
    So 2u = A + B


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    No

    A + B/(u+1)


    Not. (A+B)/(u+1)
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    Sorry to hijack this thread but are you always told what substitution you should use in exam questions?
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    (Original post by Super199)
    Sorry to hijack this thread but are you always told what substitution you should use in exam questions?
    Good Q, my friend who's done C4 said you are given the substitution... But we can never be certain with Edexcel


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    (Original post by TenOfThem)
    No

    A + B/(u+1)


    Not. (A+B)/(u+1)
    Thanks!


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    (Original post by TenOfThem)
    No

    A + B/(u+1)


    Not. (A+B)/(u+1)
    May I ask, if we had to integrate 2u-2 / u, why can we not split this into 2 and 2u^-1 ?


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    (Original post by ps1265A)
    May I ask, if we had to integrate 2u-2 / u, why can we not split this into 2 and 2u^-1 ?


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    You can

    It is possible to split a numerator but not a denominator
 
 
 
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