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    Edit: Okay nevermind I think there's nothing wrong with my method but I think I'm doing something wrong after.
    After I've intergrated it I get -1/3(x+1)^3. This is a volumes of revolution question so it's actually (1/(x+1)^2)^2 . Then I have to find the area between 1 and 0 on along the x axis rotated 360 degrees. I put in the new intergral limits 2 and 1 but I don't seem to get the right answer.
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    Just make a substitution u=1+x or so
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    Have you covered integration by substitution?
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    \int (ax+b)^n \, dx = \frac{{(ax+b)}^{(n+1)}}{a(n+1)} + k

    n \neq {-1}
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    My views on the subject of integration are that you should let whoever wants to wear a burka wear one. I know this is off-topic; I just came on this thread preparing to argue with bnp supporters. This thread is now about burkas.
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    (Original post by Necro Defain)
    Just make a substitution u=1+x or so
    Yes that's what I'm doing.
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    (Original post by Unknown?)
    After I've intergrated it I get -1/3(x+1)^3.
    Can you read my previous post? You shouldn't have a power of 3 or you need to add some more brackets because I can't follow what you are doing.
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    (Original post by Mr M)
    Can you read my previous post? You shouldn't have a power of 3 or you need to add some more brackets because I can't follow what you are doing.
    1 /3((x+1)^3)

    So I start with 1/ (x+1)^4

    sub u for x+1 so I have

    1/u^4

    make that

    u^-4

    and integrating that gives me the answer above.
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    (Original post by Mr M)
    \int (ax+b)^n \, dx = \frac{{(ax+b)}^{(n+1)}}{a(n+1)} + k

    n \neq {-1}
    What about the case a=0?
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    (Original post by Farhan.Hanif93)
    What about the case a=0?
    What about it?
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    (Original post by Mr M)
    What about it?
    Your formula crashes and burns...
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    (Original post by Unknown?)
    1 /3((x+1)^3)

    So I start with 1/ (x+1)^4

    sub u for x+1 so I have

    1/u^4

    make that

    u^-4

    and integrating that gives me the answer above.
    You have lost a sign now. Anyway, assuming that bit is correct (LaTeX would really help as you have brackets in the wrong place), have you forgotten about pi?
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    (Original post by Farhan.Hanif93)
    Your formula crashes and burns...
    Don't think so. I have been using it for about 25 years fairly happily.
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    (Original post by Mr M)
    Don't think so. I have been using it for about 25 years fairly happily.
    If a=0, your above formula has a zero in the denominator!
    All I'm saying is that there needs to be a condition up there stating that a must be non-zero. Most people will be able to spot it so I am being pretty pedantic but I guess there's not much wrong with mentioning that...
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    (Original post by Farhan.Hanif93)
    If a=0, your above formula has a zero in the denominator!
    All I'm saying is that there needs to be a condition up there stating that a must be non-zero. Most people will be able to spot it so I am being pretty pedantic but I guess there's not much wrong with mentioning that...
    Sorry I see what you mean now. Thought you were saying n = 0. Yes, accepted.
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    (Original post by Mr M)
    Sorry I see what you mean now. Thought you were saying n = 0. Yes, accepted.
    I went a bit over the top with the pedantry there...
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    (Original post by Farhan.Hanif93)
    I went a bit over the top with the pedantry there...
    No, if I had actually read what you had typed instead of what I thought you had typed all would have been fine!
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    (Original post by Mr M)
    No, if I had actually read what you had typed instead of what I thought you had typed all would have been fine!
    Not gonna argue with that. :p:
 
 
 
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