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    Okay so I did the 2014 C1 paper about an hour ago and I realised I have absolutely no clue on how to do integration and finding the area under a curve :eek3: would anyone be able to give me the run down? I've attached the question I struggled on, I think its all just knowing where to start, it really got me stumped thank you very much
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    (Original post by bethanygriffin)
    Okay so I did the 2014 C1 paper about an hour ago and I realised I have absolutely no clue on how to do integration and finding the area under a curve :eek3: would anyone be able to give me the run down? I've attached the question I struggled on, I think its all just knowing where to start, it really got me stumped thank you very much
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    What did you try?
    If you explain what you did, we might be able to point out where you went wrong - or was the whole question a bit of a mess?
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    to intergrate you just raise the power by one and then divide by the new power e.g. the intergrel of x^2 is x^3 divided by 3

    and to find the area you just intergrate the equation and then sub in the limits, so for the question you posted the limits would be -2 and 3 (they're the point A and C). you sub in 3 into the intergtaed equation then sub -2 into the intergrated equation and subtract the two
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    (Original post by chloemelissa1996)
    to intergrate you just raise the power by one and then divide by the new power e.g. the intergrel of x^2 is x^3 divided by 3

    and to find the area you just intergrate the equation and then sub in the limits, so for the question you posted the limits would be -2 and 3 (they're the point A and C). you sub in 3 into the intergtaed equation then sub -2 into the intergrated equation and subtract the two
    You might want to go back and re-read the question.

    OP, I was actually doing that paper before - where did you go wrong/ which bits do you need help with?
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    (Original post by aoxa)
    You might want to go back and re-read the question.

    OP, I was actually doing that paper before - where did you go wrong/ which bits do you need help with?
    why might i want to re-read the question, I know what to do and I know what its asking me.

    I was using x^2 as an example not taking it from the question.

    Trust me what I'm saying is correct
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    that is c2
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    (Original post by chloemelissa1996)
    why might i want to re-read the question, I know what to do and I know what its asking me.I was using x^2 as an example not taking it from the question.Trust me what I'm saying is correct
    The bit in bold was why you may have wanted to re-read the question. The limits you put were wrong for the question, which is why I told you to re-read the question.

    (Original post by indignation)
    that is c2
    No it's not. That is a C1 paper.
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    (Original post by aoxa)
    The bit in bold was why you may have wanted to re-read the question. The limits you put were wrong for the question, which is why I told you to re-read the question.



    No it's not. That is a C1 paper.
    I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
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    (Original post by Neuth)
    What did you try?
    If you explain what you did, we might be able to point out where you went wrong - or was the whole question a bit of a mess?
    The whole question was a bit of a mess :unsure: , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
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    (Original post by bethanygriffin)
    The whole question was a bit of a mess :unsure: , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from

    what do they say are the x co-ordinates for A and C?
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    (Original post by indignation)
    that is c2
    It's definitely C1 on AQA, maybe we're on different exam boards?
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    (Original post by chloemelissa1996)
    I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
    Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

    The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
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    (Original post by chloemelissa1996)
    what do they say are the x co-ordinates for A and C?
    -2 and 3
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    (Original post by bethanygriffin)
    The whole question was a bit of a mess :unsure: , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
    Did you manage to get pt ai? Showing that the x-coordinates satisfy x^2 -x-6=0?
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    (Original post by aoxa)
    Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

    The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
    Oh my gosh thank you
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    (Original post by aoxa)
    Your factorisation is correct. If you look at the question however, you're finding the shaded area R. This gives your limits to be 0 and -2 (pt c of the question does specify this 'find the area R bounded by the curve and the line segment AB' - where A is (-2,5) and B is (0,7)) Once you've found that area, you need to subtract the area below R (area of a trapezium) from the area, to leave you with area R.

    The only thing you've done wrong is misread the question - make sure to double check every question before you answer them tomorrow!
    oh yes of course! thank you
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    (Original post by bethanygriffin)
    Oh my gosh thank you
    You're welcome? I'm not sure what I've done, but if the proverbial penny has dropped I'm glad
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    (Original post by chloemelissa1996)
    I'm pretty sure the limits are x=-2 and x=3 because when you factorise x^2-x-6 you get (x+2)(x-3)=0, hence x=-2 and x=3, which are the x co-ords of A and C and hence the limits
    You are correct (not sure what the issue is here :giggle:).

    (Original post by bethanygriffin)
    The whole question was a bit of a mess :unsure: , I think if I knew how to get the coordinates of A and C I'd be able to do it. I looked on the mark scheme to try and work out what to do but I have no clue where they got their numbers from
    To start with (and find the x co-ords) you just need to factorise the equation, which gives you (x-3)(x+2)
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    (Original post by Neuth)
    You are correct (not sure what the issue is here :giggle:).
    :banghead:

    I cannot go through what limits are the correct limits discussion again... just go back, and read the question!
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    (Original post by aoxa)
    :banghead:

    I cannot go through what limits are the correct limits discussion again... just go back, and read the question!
    I never disputed that :dontknow:

    EDIT: I've just realised I quoted the wrong post... :getmecoat:
 
 
 

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