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Area Between Two Curves

  1. Aug 21, 2010 #1
    I am having trouble with this question: (QUESTION AS ATTACHED)


    This is what I tried to do:
    I found the intercepts of the difference of the two functions which I found to be only 8.

    Then I integrated this into the new function:
    [tex]\int{\sqrt{2x}-x+4}[/tex]. with limits of 0 and 8 (I don't know how to Latex this)

    This, according to my Graphics Calculator yielded 21.3.

    However, the answer is 18, what have I done wrong?

    Edit: [tex] \int_0^8 (\sqrt{2x} -x + 4)\,dx[/tex]
     

    Attached Files:

    Last edited: Aug 22, 2010
  2. jcsd
  3. Aug 21, 2010 #2

    rock.freak667

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    First, find where the line intersects the curve.

    Then from the points of intersection, draw a horizontal line to the y-axis, what shape is formed with the straight line and these two new lines you just drew? What is the area of this shape?

    EDIT:
    I see what you are doing now. Your way will be the same, you will need to subtract the same amount of times.
     
    Last edited: Aug 21, 2010
  4. Aug 21, 2010 #3

    cepheid

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    Click on my LaTeX output image below to find out:

    [tex] \int_0^8 (\sqrt{2x} -x + 4)\,dx[/tex]​
     
  5. Aug 21, 2010 #4
    I am not sure what you mean here, but is the answer a triangle?
     
  6. Aug 21, 2010 #5

    rock.freak667

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    Yep. You can get the area of the triangle. You can get the area of the region bounded by the curve and the y-axis and then subtract.
     
  7. Aug 22, 2010 #6

    hunt_mat

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    I think that this would work best as a double integral. Your y-limits are between 2 and 8 and your x-limits are between y^{2}=2x and y=x-4, this will give you the correct answer.

    Mat
     
  8. Aug 23, 2010 #7

    ehild

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    It is easier to calculate the area enclosed by the lines of the inverse functions, x=y+4 and x=y^2/2.

    ehild
     
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