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Math Help - Difficult anti differentiation question

  1. #1
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    Exclamation Difficult anti differentiation question

    I have a curve, x^3 - 14x^2 + 59x -70 , and I need to run this line through it, mx - 70 , to make the area above the curve enclosed by the line equal to HALF of the area below the line enclosed by the curve..... So, I need to find the gradient (m) of the line to run through it for this to be possible.

    So far I have derived that the x intersects are :

    x=0
    x= 7 - square root of m-10
    x= 7 + square root of m-10

    WHAT NOW
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  2. #2
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    Re: Difficult anti differentiation question

    First thing is to check on the wording of the question.
    The two areas are the same aren't they ?
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  3. #3
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    Re: Difficult anti differentiation question

    I have sent you a private msg
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  4. #4
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    Re: Difficult anti differentiation question

    So the area above the line and below the curve is \int_0^{7 - \sqrt{m-10}} (x^3 - 14x^2 + 59x -70)-(mx-70)\,dx and the area below the line and above the curve is \int_{7 - \sqrt{m-10}}^{7 + \sqrt{m-10}} (mx-70)-(x^3 - 14x^2 + 59x -70)\,dx. You need to calculate these two integrals (treating m as a constant), then set the second equal to half of the first and solve for m.

    The process is straightforward, but kind of messy.

    - Hollywood
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  5. #5
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    Re: Difficult anti differentiation question

    Quote Originally Posted by Bryanna View Post
    I have a curve, x^3 - 14x^2 + 59x -70 , and I need to run this line through it, mx - 70 , to make the area above the curve enclosed by the line equal to HALF of the area below the line enclosed by the curve..... So, I need to find the gradient (m) of the line to run through it for this to be possible.

    So far I have derived that the x intersects are :
    The x-intercepts of what? There are two graphs refered to, the graph of y= x^3- 14x^2+ 59x- 70 and y= mx- 70 and neither has what you give as x- intercepts. Do you mean where the graphs intersect, not "x-interepts"?

    x=0
    x= 7 - square root of m-10
    x= 7 + square root of m-10

    WHAT NOW
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  6. #6
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    Re: Difficult anti differentiation question

    So how would I calculate it treating m as a constant?
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    Re: Difficult anti differentiation question

    yeah I mean that
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  8. #8
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    Re: Difficult anti differentiation question

    Am I misreading this question or is everyone else ?
    '... the area above the curve enclosed by the line ...'
    '... the area below the line enclosed by the curve ...'
    They describe the same area don't they ?
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  9. #9
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    Re: Difficult anti differentiation question

    It just means pretend that m is a constant like 3 or 7. So for example,

     \int_0^{7 - \sqrt{m-10}} (x^3 - 14x^2 + 59x -70)-(mx-70)\,dx =

     \int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx =

     \left[\frac{x^4}{4} - \frac{14x^3}{3} + \frac{(59-m)x^2}{2} \right]_0^{7 - \sqrt{m-10}}

    Which will be messy if you try to calculate (7 - \sqrt{m-10})^4. Maybe you should do the other integral and hope that something cancels.

    - Hollywood
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  10. #10
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    Re: Difficult anti differentiation question

    Above the curve enclosed by the line,
    area below the curve enclosed by the line
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  11. #11
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    Re: Difficult anti differentiation question

    Difficult anti differentiation question-untitled.jpgClick image for larger version. 

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ID:	27444
    can you solve this and you will get the result of
    arctan(α/s)
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  12. #12
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    Re: Difficult anti differentiation question

    I tried it but I dont get how it would cancel out on the other side
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  13. #13
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    Re: Difficult anti differentiation question

    Here's what you should get:

    \int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx
    = -\frac{m^2}{4}+\frac{14m}{3}\sqrt{m-10}-\frac{39m}{2}-\frac{140}{3}\sqrt{m-10}+\frac{5041}{12}


    \int_{7 - \sqrt{m-10}}^{7 + \sqrt{m-10}} (mx-70)-(x^3 - 14x^2 + 59x -70)\,dx

    =\frac{28m}{3}\sqrt{m-10}-\frac{280}{3}\sqrt{m-10}

    so setting the second to half of the first gives

    -\frac{m^2}{8}-7m\sqrt{m-10}-\frac{39m}{4}+70\sqrt{m-10}+\frac{5041}{24}=0

    and the solution is

    m = 745+\frac{1274}{\sqrt{3}}-98\sqrt{112+\frac{194}{\sqrt{3}}}

    - Hollywood

    P.S. jomex2: You should start a new thread with your problem. Few (if any) people besides Bryanna and me are reading this thread now.
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  14. #14
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    Re: Difficult anti differentiation question

    ok so I am starting to see where you are coming from... But... you know how you did the first bit = - m^2/4 + 14m/3 (square root) m-10..... That answer, how did you get it? I am expanding it and I dont get how you got that
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  15. #15
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    Re: Difficult anti differentiation question

    We need powers of 7 - \sqrt{m-10}:

    (7 - \sqrt{m-10})^2 = 49 - 14\sqrt{m-10} + (m-10) = 39 + m - 14\sqrt{m-10}

    (7 - \sqrt{m-10})^3 =
    ((39 + m) - 14\sqrt{m-10} )*(7-\sqrt{m-10})=
    (273+7m)-(39+m+98)\sqrt{m-10}+14(m-10) =
    133+21m-(137+m)\sqrt{m-10}

    (7 - \sqrt{m-10})^4 =
    (133+21m-(137+m)\sqrt{m-10})*(7 - \sqrt{m-10})=
    (931+147m)-(959+7m)\sqrt{m-10})-
    (133+21m)(-\sqrt{m-10})+(137+m)(m-10)=
    (931+147m)-(1092+28m)\sqrt{m-10})+(m^2+127m-1370)=
    m^2+274m-439-(1092+28m)\sqrt{m-10})

    So the integral is:

    \int_0^{7 - \sqrt{m-10}} x^3 - 14x^2 + (59-m)x \,dx =

    \left[ \frac{1}{4}x^4 - \frac{14}{3}x^3 + \frac{59-m}{2}x^2 \right]_0^{7 - \sqrt{m-10}}=

    \frac{1}{4}(7 - \sqrt{m-10})^4 - \frac{14}{3}(7 - \sqrt{m-10})^3 + \frac{59-m}{2}(7 - \sqrt{m-10})^2 =

    \frac{1}{4}(m^2+274m-439-(1092+28m)\sqrt{m-10})) -

    \frac{14}{3}(133+21m-(137+m)\sqrt{m-10}) + \frac{59-m}{2}(39 + m - 14\sqrt{m-10}) =

    \frac{1}{4}(m^2+274m-439) -\frac{14}{3}(133+21m) + \frac{59-m}{2}(39 + m) -

    \left(\frac{1}{4}(1092+28m) - \frac{14}{3}(137+m) + \frac{59-m}{2}(14)\right)\sqrt{m-10} =

    -\frac{m^2}{4}+\frac{14m}{3}\sqrt{m-10}-\frac{39m}{2}-\frac{140}{3}\sqrt{m-10}+\frac{5041}{12}

    - Hollywood
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