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Can Someone check my work?

  1. Nov 27, 2005 #1
    I am to find the area of the region between the graphs of y=x^2 and y=-x from x=0 to x=2

    I evaluated the integral from 0 to 2 of x^2 - -x
    I got 14/3

    Is this correct?
  2. jcsd
  3. Nov 27, 2005 #2
    are you sure the function is y=-x?
  4. Nov 27, 2005 #3
    yeah, i just double checked them. y=x^2 and y= -x
  5. Nov 27, 2005 #4
    if you draw the two functions, there is no area to integrate between those curves for those limits. I am kind confused why they would say from x=0 to x=2.
    I setup the problem, and evaluated the integral with respect to y
    and got 1/6.

    i said, integral from y=1 to y = 0 and y^(.5)-y dy
  6. Nov 27, 2005 #5
    I think that they're going for the area beneath x^2 and above -x with the boundaries of x=0 and x=2.

    I know 1/6 isnt the correct answer because its not one of the choices...
    the choices are

    2/3, 8/3, 4, 14/3, and 16/3

    these are ap problems and i know they try to trick you though, so im really not sure if my answer is correct or not... i might have made up some math
  7. Nov 27, 2005 #6
    ooooh, i am sorry.. yeah then your integral ends up being x^3/3 - x^2/2
    and when you put the limits in the answer should be 2/3.
    (2)^3/3-2^2/2 = 2/3
  8. Nov 27, 2005 #7
    can you explain to me the process you used to get that?

    what i did:

    1/3*x^3 - 1/2*x^2 and then evaluated it at 2 and 0.... what did i do wrong?
  9. Nov 27, 2005 #8
    arr, stupid mistake..
    lets call f(x)=x^2 and g(x)=-x
    then integral from f(x)-g(x) = x^2+x
    integrate that function
    and we get x^3/+x^2/2
    plug in the limits
    and we get 14/3..
    sorry for the wrong answer sheesh.. i feel dumb.
  10. Nov 27, 2005 #9
    no problem. thank you so much!
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