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    (Original post by Simple Harmonic)
    It's exactly what Sean said, integration is possible when there is kx before the e^(x^2). Because to get that you differentiate e^(x^2) to give 2x e^(x^2). I hope that helps.
    Oh I see, that's why e^(x^2) stays the same. Thanks a hundred.
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    (Original post by BrainJuice)
    I was wondering why I'm left with e^{x^2} + C after integrating that. Because shouldn't e^{x^2} be integrated and turn to something else if I follow the reverse chain rule?
    No, because you can't integrate it that way.
    .
    You're 'recognising' that you've got  f'(x) \times e^{f(x)} which is the derivative of e^{f(x)} by the chain rule, hence reverse it and voila.
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    (Original post by SeanFM)
    No, because you can't integrate it that way.
    .
    You're 'recognising' that you've got  f'(x) \times e^{f(x)} which is the derivative of e^{f(x)} by the chain rule, hence reverse it and voila.
    Thank you.
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    (Original post by kkboyk)
    How do you find the height of trapezium (h) to use the trapezium rule?
    h is the 'width' of the strips.

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    (Original post by pecora)
    I also get 1/12. Which paper is this from?
    C4 January 2014 IAL
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    (Original post by SeanFM)
    h is the 'width' of the strips.

    Oh okay, so if given a table of value you just use the difference between each x value? What if you're only given the number of strips and to find the area between two x values?
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    (Original post by Music With Rocks)
    Ah so it is because it is parametric it is different, thank you very much that is perfect Just what I needed

    Thank you for sticking with me through this haha :P
    That's okay
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    (Original post by kkboyk)
    Oh okay, so if given a table of value you just use the difference between each x value? What if you're only given the number of strips and to find the area between two x values?
    Then you have to use the number of strips (given that they are all equal in 'height' (length)) to work out the height.
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    (Original post by pecora)
    I think you subtracted 2AB instead of 2BA. Have you checked with the markscheme?

    I get 33i -20j 42k
    http://qualifications.pearson.com/co...c_20130815.pdf

    pg 15 It has 17i-20j-6k

    You might need to add 2BA because they are both going into B. But because in the previous question you find AB it's easier to subtract
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    sorry guys but does anyone know the link to madasmaths for exercises on integration and vectors? can't seem to find them

    cheers!
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    (Original post by SeanFM)
    Then you have to use the number of strips (given that they are all equal in 'height' (length)) to work out the height.
    Thanks
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    and btw does anyone know what proofs or theory might come up? do we need to know surface areas of certain 3d shapes?
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    Guys do you always have to simplify direction vectors lets say line L1 has points A(1,1,1) and B(7,13,25)

    so is the direction vector (6,12,24) or (1,2,4) ?????

    please clearify
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    has anyone done the june 2016 C34 IAL paper yet - this exam was 2 days ago? The C3 questions in it were very similar to the ones that we had on our paper. Maybe with a bit of luck we may have similar C4 questions that were on that paper..
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    https://drive.google.com/file/d/0B1Z...w?pref=2&pli=1
    june 2016 IAL
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    (Original post by Gitty)
    has anyone done the june 2016 C34 IAL paper yet - this exam was 2 days ago? The C3 questions in it were very similar to the ones that we had on our paper. Maybe with a bit of luck we may have similar C4 questions that were on that paper..
    https://drive.google.com/file/d/0B1Z...w?pref=2&pli=1
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    guys you know when they ask you to 'use your approximation to estimate an approximate value of blah blah and you have to find X. how do you find x.? for example, ques 4b here:

    http://qualifications.pearson.com/co...e_20130618.pdf
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    (Original post by M&F)
    Guys do you always have to simplify direction vectors lets say line L1 has points A(1,1,1) and B(7,13,25)

    so is the direction vector (6,12,24) or (1,2,4) ?????

    please clearify
    No you wouldn't have to! But it would be easier to do algebra with (1,2,4) rather than (6,12,24) - it's just the value of  \lambda will be different in  r = a + \lambda b
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    (Original post by Gitty)
    has anyone done the june 2016 C34 IAL paper yet - this exam was 2 days ago? The C3 questions in it were very similar to the ones that we had on our paper. Maybe with a bit of luck we may have similar C4 questions that were on that paper..
    yeah i saw it, the C4 parts to it were piss easy. I doubt we'll get anything similar
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    (Original post by M&F)
    But in some questions I have seen it being simplified and not simplifying gives a different answer like Q7 on this paper.

    http://www.madasmaths.com/archive/iy...apers/c4_m.pdf .
    I haven't looked at the question but the direction vector is preceded by a constant e.g. K(6,12,24) so it can be simplified to k(1,2,4) because k is an arbitrary constant (it can take on any value).
 
 
 
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