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    (Original post by kiiten)
    Ahh i see - i remember this method now. Do you get x^3 ?
    thats perfect

    now what are your thinking for the  \displaystyle \frac{32}{x^2} term?
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    (Original post by DylanJ42)
    thats perfect

    now what are your thinking for the  \displaystyle \frac{32}{x^2} term?
    so that would be 32x-2 which differentiates to - 64x^-3 ?
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    (Original post by kiiten)
    so that would be 32x-2 which differentiates to - 64x^-3 ?
    Perfect.
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    (Original post by kiiten)
    so that would be 32x-2 which differentiates to - 64x^-3 ?
    nicely done
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    Quick question. For 8.c) is it 16 - 40/3 = 2.67
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    (Original post by kiiten)
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    its a full solution zack, but at least it's a colourful full solution
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    (Original post by DylanJ42)
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    its a full solution zack, but at least it's a colourful full solution
    Hmm i thought there might be a x2 somewhere. So, do you do 16 - 8/3 - 8/3 ?
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    (Original post by kiiten)
    Hmm i thought there might be a x2 somewhere. So, do you do 16 - 8/3 - 8/3 ?
    you would indeed
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    (Original post by DylanJ42)
    you would indeed
    Thank you!! - Before i was doing 16 - 40/3 - 40/3 so i kept getting a -ve answer. I see where ive gone wrong now. Thanks
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    Thanks for all your help everyone, I appreciate it
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    Different ques - 13.a) I got x^4 = 9 Does that mean there will be a +ve & -ve answer?

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    (Original post by kiiten)
    Different ques - 13.a) I got x^4 = 9 Does that mean there will be a +ve & -ve answer?

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    it does indeed
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    (Original post by DylanJ42)
    it does indeed
    Ah, thanks :yes:

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    (Original post by DylanJ42)
    it does indeed
    Wait what about the shaded area? You have to prove its 5 1/3.
    I integrated using the bounds 3 and 1 but got an area of 25. I think you have to use the y =4 part but i dont know how? :s

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    (Original post by kiiten)
    Wait what about the shaded area? You have to prove its 5 1/3.
    I integrated using the bounds 3 and 1 but got an area of 25. I think you have to use the y =4 part but i dont know how? :s

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    When you integrate between 1 and 3 you are finding the area enclosed between the curve, the x axis, x=1 and x=3 (look at the photo, area = part surrounded in green highlighter)

    so you need to find area then subtract the area of the rectangle formed under the shaded area

    Also check your integration, the integral shouldn't be 25

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    (Original post by DylanJ42)
    its a full solution zack, but at least it's a colourful full solution
    Not even mad. Think this is an appropriate situation for a full solution. :yep:
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    (Original post by DylanJ42)
    When you integrate between 1 and 3 you are finding the area enclosed between the curve, the x axis, x=1 and x=3 (look at the photo, area = part surrounded in green highlighter)

    so you need to find area then subtract the area of the rectangle formed under the shaded area

    Also check your integration, the integral shouldn't be 25

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    So i integrated 14 - x^2 - 9x^-2 to get 14x - 1/3x^3 + 9x^-1 is that right so far?
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    (Original post by kiiten)
    So i integrated 14 - x^2 - 9x^-2 to get 14x - 1/3x^3 + 9x^-1 is that right so far?
    yep, all good so far
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    (Original post by DylanJ42)
    yep, all good so far
    Then i subbed in 3 and 1 separately then took them away to get 25? :dontknow:

    So 14 x 3 - 1/3 × 3^2 + 9 × 3^-1 for the first term (3)
 
 
 
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