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Math Help - Proper Integration

  1. #1
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    Proper Integration

    Hello I am new.

    Question:
    Find the area bounded by the graph of each of the following functions, the x-axis and the given co-ordinates. Sketch the graph in case.
    f(x)=2-x-x^2 ; x=0 to x=2.

    My attempt:
    =2x-(x^2/2)-(x^3/3)
    =(0) - (2(2) - (2)^2/2) - (2)^3/3)
    =0- (4-2-(8/3))
    =0 - (-2/3)
    Absolute value gives me:
    2/3
    But the answer is not right it is:
    3.
    I don't get it please help.
    Thank you.
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  2. #2
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    Is the answer right?
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  3. #3
    MHF Contributor

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    The graph of y= 2- x- x^2 crosses the x-axis at x= 1. The region bounded by those graphs has a portion above the x-axis and a portion below it. Since area is always positive and integrating a negative function give a negative result the area is
    \int_0^1 (2- x- x^2) dx+ \int_1^2 -(2- x- x^2) dx
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