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can you integrate like this?

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    y=x+1

    I(y)=((x+1)^2)/2)
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    (Original post by KloppOClock)
    y=x+1

    I(y)=((x+1)^2)/2)
    Yeah you can but when you do it that way you will get a different value for the constant than if you did it the normal way.
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    (Original post by rayquaza17)
    Yeah you can but when you do it that way you will get a different value for the constant than if you did it the normal way.
    okay so that works always for definite integration. is the method just to square the entire polynomial and divide by 2? I just saw this in a mark scheme and its the first time ive seen it done this way
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    (Original post by KloppOClock)
    okay so that works always for definite integration. is the method just to square the entire polynomial and divide by 2? I just saw this in a mark scheme and its the first time ive seen it done this way
    Was it on mat.
    I remember


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    (Original post by KloppOClock)
    okay so that works always for definite integration. is the method just to square the entire polynomial and divide by 2? I just saw this in a mark scheme and its the first time ive seen it done this way
    Let u=x+1 it follows.
    It works in general for ax+b but not for higher powers obviously.


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    (Original post by KloppOClock)
    okay so that works always for definite integration. is the method just to square the entire polynomial and divide by 2? I just saw this in a mark scheme and its the first time ive seen it done this way
    Like physicsmaths says it is using the method of how you would integrate (ax+b)^n.
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    (Original post by physicsmaths)
    Let u=x+1 it follows.
    It works in general for ax+b but not for higher powers obviously.
    Note that \displaystyle \int (ax+b) \,dx = \dfrac{1}{a} \dfrac{(ax+b)^2}{2} + C, that is, you also need to divide by a.

    (I know you know this, but the simplistic rule the OP gave doesn't hold for a \neq 1).
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    (Original post by DFranklin)
    Note that \displaystyle \int (ax+b) \,dx = \dfrac{1}{a} \dfrac{(ax+b)^2}{2} + C, that is, you also need to divide by a.

    (I know you know this, but the simplistic rule the OP gave doesn't hold for a \neq 1).
    Ah yes, Sorry I wasn't clear about the recognition part(dividing by a etc!)
 
 
 
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