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Edexcel C3,C4 June 2013 Thread Watch

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    Bring on vectorgeddon
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    Has anyone come across an integration by inspection on past papers? (Chapter 6E)
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    (Original post by justinawe)
    You can do it either way.
    you can?????????????????? does it work for every type of solution????
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    Anyone know any hard vector questions. Also, for vector questions, when are we allowed to let x and y equal certain value? much appreciated
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    (Original post by Zaphod77)
    No. You can only use the ln rule when the top of the fraction is the differential of the bottom, or a scalar multiple of it. u is not a scalar as it is not a number, it is a function. To integrate that you'd need to use long division or the methods mentioned earlier to separate it into stuff you can integrate
    Oh right.. So you wouldnt be able to split it like u/4 + u/u ?
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    (Original post by justinawe)
    You don't need to divide through for equal powers of x.

    The only difference here would be that, instead of splitting it into:

    \dfrac{A}{2x+1} + \dfrac{B}{2x-1}

    You'd split it into:

    \dfrac{A}{2x+1} + \dfrac{B}{2x-1} + \dfrac{C}{(2x+1)(2x-1)}
    Ah, I was never taught that, thank you I think I'll stick to dividing though, just because that's what I've always done!
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    How do you integrate 2^x
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    You're great
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    (Original post by Story)
    Oh right.. So you wouldnt be able to split it like u/4 + u/u ?
    If it was (u+4)/u then you could split it into u/u + 4/u, but you can't split the denominator up like that - think when you add fractions, you can't add them unless the denominator is the same
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    (Original post by Kreayshawn)
    what paper is this from?
    Differentiation solomon worksheet F
    Its not a solomon paper

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    Good luck in the C4 exam tomorrow guys, I hope you all have been revising S2
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    (Original post by Zaphod77)
    If it was (u+4)/u then you could split it into u/u + 4/u, but you can't split the denominator up like that - think when you add fractions, you can't add them unless the denominator is the same
    Thank you! I still dont understand how you would split u/u+4 I tried doing (u+4) (something) = U ....

    Is the something -4?...How would you form it into a fraction then?

    By the way this is for the question integrate 1/4+root x+1 .. where u = root x+1.

    I just get confused a little with some substitution q's.
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    (Original post by subturfuge)
    Good luck in the C4 exam tomorrow guys, I hope you all have been revising S2
    I took a good look at FP3 instead
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    (Original post by Ender'sgame)
    How do you integrate 2^x
    I'm assuming you know that y = 2^x \ \Rightarrow \ \dfrac{dy}{dx} = 2^x \ln 2

    So,

    \displaystyle \int 2^x \ dx

    = \displaystyle \frac{1}{\ln 2} \int 2^x \ln 2 \ dx

    = \dfrac{2^x}{ \ln 2} + C
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    (Original post by Story)
    Thank you! I still dont understand how you would split u/u+4 I tried doing (u+4) (something) = U ....

    Is the something -4?...How would you form it into a fraction then?
    U/(U+4) = (U-4+4)/(U+4) does this help now?
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    (Original post by justinawe)
    I'm assuming you know that y = 2^x \ \Rightarrow \ \dfrac{dy}{dx} = 2^x \ln 2

    So,

    \displaystyle \int 2^x \ dx

    = \displaystyle \frac{1}{\ln 2} \int 2^x \ln 2 \ dx

    = \dfrac{2^x}{ \ln 2} + C
    Thank you.
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    (Original post by Story)
    Thank you! I still dont understand how you would split u/u+4 I tried doing (u+4) (something) = U ....

    Is the something -4?...How would you form it into a fraction then?

    By the way this is for the question integrate 1/4+root x+1 .. where u = root x+1.

    I just get confused a little with some substitution q's.
    Ok, there are 2 methods you could use. 1 is that you make the top u+4-4, so then you could split the fractions into (u+4)/(u+4) - 4/(u+4), which you can then integrate, as the first fraction simplifies down to equal 1. The other method is dividing u by u+4, which gives you a whole number and a remainder which you can then integrate
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    Hey guys!

    Can someone help me with Edexcel C4 review exercise question 74 (c)? I don't get why we have to integrate from 0.5 to 2....Thanks in advance
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    (Original post by justinawe)
    I'm assuming you know that y = 2^x \ \Rightarrow \ \dfrac{dy}{dx} = 2^x \ln 2

    So,

    \displaystyle \int 2^x \ dx

    = \displaystyle \frac{1}{\ln 2} \int 2^x \ln 2 \ dx

    = \dfrac{2^x}{ \ln 2} + C
    am I right I thinking when you differentiate you multiply by ln(2) and when you integrate you divide by ln(2)
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    (Original post by MathsNerd1)
    U/(U+4) = (U-4+4)/(U+4) does this help now?
    Yes, but how would it split all the way?
 
 
 
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