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Math Help - How is this problem incorrect? Slope with Integration

  1. #1
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    How is this problem incorrect? Slope with Integration

    There is a line through the origin that divides the region bounded by the parabola y=8x−4x2 and the x-axis into two regions with equal area. What is the slope of that line?

    Well I integrated the fuction from zero to 2, and got 16/3. Took half the area. Integrated with mx, and got 2m. Then set it equal to 16/6 and solving for m found the slope to be 4/3. My online hw says this is incorrect. How? and what's the actual answer?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by radioheadfan View Post
    There is a line through the origin that divides the region bounded by the parabola y=8x−4x2 and the x-axis into two regions with equal area. What is the slope of that line?

    Well I integrated the fuction from zero to 2, and got 16/3. Took half the area. Integrated with mx, and got 2m. Then set it equal to 16/6 and solving for m found the slope to be 4/3. My online hw says this is incorrect. How? and what's the actual answer?
    These two graphs intersect at x=2-\frac{m}{4}

    So for the two areas to be equal,

    \int_0^{2-m/4}(8x-4x^2-mx)\,dx=\int_0^{2-m/4}mx\,dx+\int_{2-m/4}^2(8x-4x^2)\,dx

    (Try to visualize geometrically why this is true.)

    Do some number crunching and solve for m.
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  3. #3
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    Quote Originally Posted by radioheadfan View Post
    There is a line through the origin that divides the region bounded by the parabola y=8x−4x2 and the x-axis into two regions with equal area. What is the slope of that line?

    Well I integrated the fuction from zero to 2, and got 16/3. Took half the area. Integrated with mx, and got 2m.

    You integrated mx...between what and what? This doesn't sound correct: it should have been between 0 and 2 -m/4 = the x-coordinate of the intersection point of the parabola and the line, and then you still need to add to this the integral from 2 - m/4 to 2 of the parabola!
    This is a very nasty exercise, and the best I got is an awful cubic equation in m.

    Tonio


    Then set it equal to 16/6 and solving for m found the slope to be 4/3. My online hw says this is incorrect. How? and what's the actual answer?
    .
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  4. #4
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    I tried doing this problem out again for like the tenth time and I still can't get the right answer. Can someone just help me do this. I'm sick of this problem.
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  5. #5
    Super Member redsoxfan325's Avatar
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    I just told you exactly how to do it! I'm not going to actually calculate it for you.
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