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    My aunt gave me this oldschool textbook thats older than me to work out of, and its alot harder than what im used to.
    I cant integrate these two functions

    \int (2+3x)^6\ dx

    With upper limit 0 , lower limit -1.

    and


    \int (4x-1)^1/3\ dx

    upper limit 7, lower limit 1/2

    Hehe sorry for my newb use of latex, on the last example the whole bracket is to the power of 1/3. I know the answers i just.. cant get them ..
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    In general, the integral of (ax +b)^n = (1/a(n+1))*(ax+b)^(n+1) for any n not equal to -1. This is from C4.
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    (Original post by vc94)
    In general, the integral of (ax +b)^n = (1/a(n+1))*(ax+b)^(n+1) for any n not equal to -1. This is from C4.
    C4 WTF. Well im on C2 and she was like 'yeah it will help you with your exam!'
    lol...
    cheers i will try that and see where i end up
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    make a substitution u = 2+3x
    change so you are integrating u w.r.t. u
    du/dx = 3, so dx=du/3
    change the limits, so
    u=2+3(0) = 2
    u= 2+3(-1) = -1
    gives you upper limit 2, lower limit -1

    giving you the integral
    \int^2_{-1} u^6 \frac{du}{3}

    which should be fairly straight forward.

    do the second one in the same way.

    I think all that is right. I'm pretty tired.
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    (Original post by TimmonaPortella)
    make a substitution u = 2+3x
    change so you are integrating u w.r.t. u
    du/dx = 3, so dx=du/3
    change the limits, so
    u=2+3(0) = 2
    u= 2+3(-1) = -1
    gives you upper limit 2, lower limit -1

    giving you the integral
    \int^2_{-1} u^6 \frac{du}{3}

    which should be fairly straight forward.

    do the second one in the same way.

    I think all that is right. I'm pretty tired.
    Thank alot! Im only on C2 and didn't realise i'd be going so off track!
    This core mathematics book lied to me when it didnt tell me it was going to make me do stuff i didn't know
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    (Original post by Jampolo)
    Thank alot! Im only on C2 and didn't realise i'd be going so off track!
    This core mathematics book lied to me when it didnt tell me it was going to make me do stuff i didn't know
    Well you can just use inspection, as the other guy said.
    You won't really understand why that works without understanding substitution, though.

     \int (ax+b)^n dx
     u=ax+b \rightarrow \frac{du}{a}=dx
    Subbing in U and swapping dx for du/a gives
     \int \frac{u^n}{a} du
    which integrates to  \frac{u^{n+1}}{a(n+1)} + c
    Which is the formula the other poster gave, when you sub back in your f(x) for u

    This is all entirely unnecessary for core 2, though.
    Not really sure why your teacher said they would help for that exam.

    Anyway, hope some of that helped with whatever objective you're going for
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    this is basic c4, (1/((n+1)a)) (ax+b)^(n+1)) +c

    I know this probably has already been said but I wanted to type it out
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    (Original post by TimmonaPortella)
    Well you can just use inspection, as the other guy said.
    You won't really understand why that works without understanding substitution, though.

     \int (ax+b)^n dx
     u=ax+b \rightarrow \frac{du}{a}=dx
    Subbing in U and swapping dx for du/a gives
     \int \frac{u^n}{a} du
    which integrates to  \frac{u^{n+1}}{a(n+1)} + c
    Which is the formula the other poster gave, when you sub back in your f(x) for u

    This is all entirely unnecessary for core 2, though.
    Not really sure why your teacher said they would help for that exam.

    Anyway, hope some of that helped with whatever objective you're going for
    I sort of understand it, this book give sthe same formula. Its just that my aunt gave me loads of books and papers to work on, but they're just titled 'Core Mathematics' and 'Pure Mathematics' so im not really sure what module im actually working on

    And yeah it did help thanks, made me realise im attempting i probably wont be for almost another year
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    (Original post by gunmetalpanda)
    this is basic c4, (1/((n+1)a)) (ax+b)^(n+1)) +c

    I know this probably has already been said but I wanted to type it out
    u mad?
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    (Original post by Jampolo)
    u mad?
    Huh? And I think you added me on skype
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    (Original post by gunmetalpanda)
    Huh? And I think you added me on skype
    your sig
    And ooh whats your user?
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    (Original post by Jampolo)
    My aunt gave me this oldschool textbook thats older than me to work out of, and its alot harder than what im used to.
    I cant integrate these two functions

    \int (2+3x)^6\ dx

    With upper limit 0 , lower limit -1.

    and


    \int (4x-1)^1/3\ dx

    upper limit 7, lower limit 1/2

    Hehe sorry for my newb use of latex, on the last example the whole bracket is to the power of 1/3. I know the answers i just.. cant get them ..
    Like others have said, these are C4. If you're doing Edexcel AS Maths, C2 integration is identical in difficulty to C1 integration but with limits and areas under curves.
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    (Original post by Salliana23)
    Like others have said, these are C4. If you're doing Edexcel AS Maths, C2 integration is identical in difficulty to C1 integration but with limits and areas under curves.
    Im ocr where we dont even meet integration until C2!
 
 
 
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