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Math Help - Definite Integration

  1. #1
    gp3
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    Definite Integration

    hi guys, i dont know where to place this integration question under, so i guess i'lljust place it here. I cant get the answer after many tries..

    Find the area bounded by the given curves:

    y=4-x square ; y=0, x=0, x=4

    the answer given to me was 16. but I could only get -15.99 yes the ans I got is negative.

    i solved by coming out with the table, and then integrate the equation by sub in the limit 4 and 0.

    x 0 1 2 3 4
    y 4 3 0 -5 -12

    anyone help pls? thanks a lot! I have end of semester exam coming!
    Last edited by gp3; February 9th 2010 at 09:51 PM.
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  2. #2
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    Hello, gp3!

    Find the area bounded by: . y\,=\,4-x^2,\;y=0,\;x=0,\;x=4

    The answer given to me was 16.

    I believe the area looks like this:
    Code:
                      |
                     4*..
                  *   |:::*.      4
          -----*------+------*----+----
             *        |      2 *:::
            *         |         *:: 
                      |         ':: 
           *          |          *: 
                      |          ': 
                      |           : 
          *           |           * 
                      |

    We want the area above the x-axis from 0 to 2:

    . . \int^2_0(4-x^2)\,dx \;\;=\;\;4x - \tfrac{1}{3}x^3\,\bigg]^2_0  \;\;=\;\;\left(8 - \tfrac{8}{3}\right) - (0-0) \;\;=\;\;\frac{16}{3}


    We want the area below the x-axis from 2 to 4:

    . . \int^4_2(x^2-4)\,dx \;\;=\;\;\tfrac{1}{3}x^3-4x\,\bigg]^4_2 \;\;=\;\;\left(\tfrac{64}{3} - 32\right) - \left(\tfrac{8}{3} - 8\right) \;\;=\;\;\tfrac{16}{3} - \left(-\tfrac{16}{3}\right) \;\;=\;\;\frac{32}{3}


    Total area: . \frac{16}{3} + \frac{32}{3} \;\;=\;\;\frac{48}{3} \;\;=\;\;16

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  3. #3
    gp3
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    hi thanks a lot!

    but one thing i don't understand is that why the question have to give me y=0? I thought we just need to know the range of x axis which is 0 to 4.
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