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Math Help - Calc. Problem

  1. #1
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    Calc. Problem

    Calculate the area under the graph of f(x) = 3x^2 and above the x-axis on the interval from x= -2 to x= 2.

    I entered this graph into my calculator and there is no area under the graph, so I assumed it was 0. Then, I did fnInt(3x^2,x,-2,2) and got 16 as my solution. Just was wondering if I could get a second opinion to see if I did this right.
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  2. #2
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    Quote Originally Posted by redsoxfan0825 View Post
    Calculate the area under the graph of f(x) = 3x2 and above the x-axis on the interval from x= -2 to x= 2.

    I entered this graph into my calculator and there is no area under the graph, so I assumed it was 0. Then, I did fnInt(3x2,x,-2,2) and got 16 as my solution. Just was wondering if I could get a second opinion to see if I did this right.
    The graph of 3x^2 is symmetrical about x=0 so we can consider one side of x=0 and multiply by 2

    \int^2_{-2}3x^2\,.dx = 2\, \int^2_0 3x^2\,.dx

    That integral should be simple enough and equal 8 so when we multiply that by 2 we get 16

    Not sure how you got 0 either because integrating the original expression yields 8-(-8)=16

    edit: attached is the graph of 3x^2 and it clearly shows an area
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