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Workshop question

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the parabola y=a(x^2) that divides the area under the curve y=x(1-x) over [0,1] into two regions of equal area.

    2. Relevant equations
    I set the two equations equal to each other to solve that the intersection point is x=1/(1+a).
    I solved for the entire area "definite integral x(1-x) from [0,1] dx" = 1/6.

    3. The attempt at a solution
    I attempt half the area with the two definite integrals "a(x^2) from [0,(1/(1+a))]" + the integral "x(1-x) from [(1/(1+a)), 1]" set equal to 1/12 (half the area). But I can not solve for a.

    It looks like there is Linear algebra but I am only in a Calc II class so it should not be too hard. Please help.
     
  2. jcsd
  3. Sep 14, 2008 #2

    Dick

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    The area under y=x(1-x) is supposed to split into two regions of area 1/12. One of them is the region bounded by y=x(1-x) above and y=ax^2 below between 0 and 1/(1+a). Figure out what that is and set it equal to 1/12. The other region will take care of itself.
     
  4. Sep 14, 2008 #3
    That is what I have done. But I come up with an equation too hard to solve for a. I believe my problem lies in my algebra.
     
  5. Sep 14, 2008 #4

    Dick

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    Ok. So what do you get for the integral in terms of a? I get an equation that looks like it might be a cubic, but when you do the algebra the cubic term cancels out and it becomes a quadratic.
     
  6. Sep 14, 2008 #5
    That is what I thought I was doing wrong but I have an addition of the cubics. My integral is x-x^2 from 0 to (1/1+a) - ax^2 from x to (1/1+a).
     
  7. Sep 14, 2008 #6

    Dick

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    Right so far except the second limit is also 0 to 1/(1+a), just a typo I hope. So put the limits in and subtract. Bring everything to a common denominator and set it equal to 1/6. The cubic terms canceled for me. If you show us the intermediate steps maybe we can see what went wrong.
     
  8. Sep 14, 2008 #7
    Yes sorry that was a typo. After integration I get (x^2/x)-(x^3/3)-a(x^3/3). I set this equal to 1/12 but as you see the cubic terms do not cancel.
     
  9. Sep 14, 2008 #8

    Dick

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    Put x=1/(1+a) and THEN try and do the algebra. The cubic parts in a will cancel.
     
  10. Sep 14, 2008 #9
    I'm not sure I understand. After I integrate substitute x for (1/1+a). If I do this I get "((1/1+a)^2/2)-((1/1+a)^3/3)-a((1/1+a)^3/3) and they still do not cancel.
     
  11. Sep 14, 2008 #10

    Dick

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    You haven't done any simplification yet. I told you, bring them to a common denominator and sum them all into one term.
     
  12. Sep 14, 2008 #11
    I'm confused. Now I have (1/1+a)^2(3-2(1/1+a)-2a(1/1+a))=1/2. I still have the cubic terms.
     
  13. Sep 14, 2008 #12

    Dick

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    You aren't done yet! Add up the terms in the parentheses!
     
  14. Sep 14, 2008 #13
    Yes I did that but I don't get it, there is still a cubic.
     
  15. Sep 14, 2008 #14

    Dick

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    Argh. What is 3-2/(1+a)-2a/(1+a)? Common denominator. Add them.
     
  16. Sep 14, 2008 #15
    Common denominator is (1+a) so I have ((3(1+a)-2-2a)/(1+a))(1/(1+a))^2. This would give me (3(1+a)-2-2a)(1/(1+a))^3=1/2. I am so frustrated. I don't get it!
     
  17. Sep 14, 2008 #16

    Dick

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    You are doing everything right but you keep taking one step and then stopping. 3(1+a)-2-2a=3+3a-2a-2=1+a. Do you see the cancellation now??
     
  18. Sep 14, 2008 #17
    Wow, thank you so much, I don't know why I couldn't see that. a=0.41421. Thanks again.
     
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