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    I am stuck on the 3rd question.

    y=x^2-5x+4. I then factorised that to (x+5) and (x-1)

    Therefore; x=-5 and x=1

    I integrated to get this:


    I thought that I would need to find the range of (0; -5) (1, 5) (1,0) and (1, -5)

    Not 100% sure though.
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    (Original post by apronedsamurai)
    I am stuck on the 3rd question.

    y=x^2-5x+4. I then factorised that to (x+5) and (x-1)

    Therefore; x=-5 and x=1

    I integrated to get this:


    I thought that I would need to find the range of (0; -5) (1, 5) (1,0) and (1, -5)

    Not 100% sure though.
    Your factorisation is wrong:

    \displaystyle (x+5)(x-1)\neq x^2-5x+4
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    (Original post by notnek)
    Your factorisation is wrong:

    \displaystyle (x+5)(x-1)\neq x^2-5x+4
    Ok, (x-4)(x-1)

    x=4; x=1
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    (Original post by apronedsamurai)
    Ok, (x-4)(x-1)

    x=4; x=1
    Do you still need help?
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    Sadly, yes.

    I thought that I would get the area where, 4 and 1 where the integers.

    using the integrated forms, my final answer was (-) 13/6; (I have put brackets around the negative sign to denote the fact that I am aware that you disregard the negatives with the areas).

    There are a total of three answers given: 1,4 and 6 to a 1/3

    :s

    Am now utterly lost :/
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    Where did they get three values from?
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    (Original post by apronedsamurai)
    Sadly, yes.

    I thought that I would get the area where, 4 and 1 where the integers.

    using the integrated forms, my final answer was (-) 13/6; (I have put brackets around the negative sign to denote the fact that I am aware that you disregard the negatives with the areas).

    There are a total of three answers given: 1,4 and 6 to a 1/3

    :s

    Am now utterly lost :/
    I think I may know what you're doing wrong but your answer is not what I expected.

    So can you post your working?

    I'm not sure what you mean by "3 answers given". Do you mean for parts a), b) and c)? The correct answer for c) is 6\frac{1}{3}.
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    The answer section of the book gives threes results/answers:

    1, 4 and 6 1/3

    Ok my working.





    =-8/3


    = - 1/2


    =
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    (Original post by apronedsamurai)
    The answer section of the book gives threes results/answers:

    1, 4 and 6 1/3
    1 and 4 are where the graph cuts the x-axis. That's the first part of the question.

    The second part asks you to find the area and that answer is 6\frac{1}{3}.
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    No, I am focusing solely and exclusively on c).

    For c) the answers listed for that question are:

    1, 4 and 6 1/3
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    (Original post by apronedsamurai)
    No, I am focusing solely and exclusively on c).

    For c) the answers listed for that question are:

    1, 4 and 6 1/3
    Read my last post.
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    Sorry that was a delayed response, in response to one of your previous posts.



    Ok; so in essence then I had already identified 2 of the 3 answers.

    So where am I going wrong, with regards to getting 6 1/3?
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    (Original post by apronedsamurai)
    The answer section of the book gives threes results/answers:

    1, 4 and 6 1/3

    Ok my working.





    =-8/3


    = - 1/2


    =
    You're using \displaystyle \frac{x^3}{3}-\frac{5x^2}{2}-4x instead of \displaystyle \frac{x^3}{3}-\frac{5x^2}{2}+4x.

    The latter is the correct version, which you gave in your initial post.
 
 
 
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