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Area bounded between 2 graphs

  1. Jul 23, 2012 #1
    1. The problem statement, all variables and given/known data

    f(x)=x2-1 and f(x)=2x+2

    2. Relevant equations



    3. The attempt at a solution

    Points of intersection are -1 and 3. So you integrate using those as upper and lower and plug it in and subtract, right? But I get 0 for each. So nothing to subtract and 0 is not the correct answer.
     
  2. jcsd
  3. Jul 23, 2012 #2

    tiny-tim

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    Hi XodoX! :smile:
    yes :smile:
    how? :confused:

    show us your integrations​
     
  4. Jul 23, 2012 #3
    Well, you set both to 0 and basically combine them, then you get x^2 - 2x - 3. If you plug in -1 and 3 for it, then you get 0.
     
  5. Jul 23, 2012 #4
    So at what point did you do any integration?
     
  6. Jul 23, 2012 #5

    tiny-tim

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    but you haven't integrated!! :rolleyes:

    all you've done is find the points where their difference is 0​

    go forth and integrate!! :smile:
     
  7. Jul 23, 2012 #6
    Never-mind. Wrong number.

    I get -10.6 after integrating the combined equation. Plug in 3 and subtract it from what I get for -1.

    x3/3 - x2-3x

    No, it's +10.6. Sorry.
     
  8. Jul 23, 2012 #7

    Mark44

    Staff: Mentor

    An area should be nonnegative. If you got a negative number, your integrand is set up incorrectly.

    Note that the line is above the parabola throughout the interval.
     
  9. Jul 23, 2012 #8
    Yes, I did. I did the 3 first and then the -1. I thought that's how you did that.
     
  10. Jul 23, 2012 #9

    Mark44

    Staff: Mentor

    I'm talking about the integrand, not the limits of integration. I believe you set up the integral as:
    $$ \int_{-1}^3 (x^2 - 1) - (2x + 2)~dx$$

    That will give you a negative number.
     
  11. Jul 24, 2012 #10

    HallsofIvy

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    Because you have them "wrong way around". [itex]2x+2> x^2- 1[/itex] for all x between -1 and 3.
     
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