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Math Help - A negative area?? am i doing this wrong

  1. #1
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    A negative area?? am i doing this wrong

    who likes my paint skills? lol



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  2. #2
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    Have you considered learning to use LaTeX?
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  3. #3
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    Quote Originally Posted by Legendsn3verdie View Post
    who likes my paint skills? lol



    Use symmetry. The enclosed area is twice the area from x = 0 to x = 2:


    Enclosed area = - 2 \int_{0}^{\sqrt{2}} (x^2 - 2) \, dx + 2 \left( 4 - \int_{\sqrt{2}}^{2} (x^2 - 2) \, dx \right).


    The negative ensures you get a positive number for that section of the area. The 4 is the area of the obvious square.
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    Quote Originally Posted by mr fantastic View Post
    Use symmetry. The enclosed area is twice the area from x = 0 to x = 2:


    Enclosed area = - 2 \int_{0}^{\sqrt{2}} (x^2 - 2) \, dx + 2 \left( 4 - \int_{\sqrt{2}}^{2} (x^2 - 2) \, dx \right).


    The negative ensures you get a positive number for that section of the area. The 4 is the area of the obvious square.
    way to advanced for me no idea what you did
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    Super Member 11rdc11's Avatar
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    What part didn't you understand?
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    Quote Originally Posted by 11rdc11 View Post
    What part didn't you understand?

    how to get the area of the full curve.. did i do it right?
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    Quote Originally Posted by Legendsn3verdie View Post
    how to get the area of the full curve.. did i do it right?
    no.

    A = \int_{-2}^2 2 - (x^2 - 2) \, dx = 2 \int_0^2 4 - x^2 \, dx = \frac{32}{3}<br />
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  8. #8
    Super Member 11rdc11's Avatar
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    Mr. Fantastic just the broke the integral in half because everything below the x axis would be negative. So he found the area below and then added it to area above the x axis from x = 0 to x = 2. That would give you the area in the 1st and 4th quad. Now since this is an even function multiply the area from quad 1 and 4 by 2
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