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Math Help - Finding area of region bounded (Integration)

  1. #1
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    Finding area of region bounded (Integration)

    Hi, so I have this problem:
    To find the area of the region bounded by graphs of y=x3, y=0, x=1, and x=3 by limit definition

    But I was a bit confused on this because I didn't really know what the intervals were.
    Like I graphed it out, and it would look like this:
    I will describe it: A horizontal line on y=0, then 2 veritcal lines on x=1 and x=3, then the x3 curves from down to up, hits the 0, but the shape looks pretty weird.

    Can someone explain to me what area/intervals are supposed to be?
    From my intepretation when I saw the graph of all those equations, it looked like it would be [1, 3] since 3 looks like the max, and 1 is like the min, and there's where they all come together

    They make somewhat like a rectangle at the bottom, then it curves up, so like a curvy trapezoid? Maybe?
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  2. #2
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    Re: Finding area of region bounded (Integration)

    I think you have the region right - its boundaries are the vertical lines x=1 and x=3 on the right and left, the horizontal line y=0 on the bottom, and the curve y=x^3 on the top.

    The way I learned it was to divide the region into vertical rectangles. These rectangles have width dx and height x^3, so the sum of x^3\,dx is the area of the region. In the limit, it's the integral of x^3\,dx and the limits are x=1 on the left and x=3 on the right. So the area is given by:

    \int_1^3x^3\,dx.

    Hopefully that answers your question.

    - Hollywood
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  3. #3
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    Re: Finding area of region bounded (Integration)

    Quote Originally Posted by hollywood View Post
    I think you have the region right - its boundaries are the vertical lines x=1 and x=3 on the right and left, the horizontal line y=0 on the bottom, and the curve y=x^3 on the top.

    The way I learned it was to divide the region into vertical rectangles. These rectangles have width dx and height x^3, so the sum of x^3\,dx is the area of the region. In the limit, it's the integral of x^3\,dx and the limits are x=1 on the left and x=3 on the right. So the area is given by:

    \int_1^3x^3\,dx.

    Hopefully that answers your question.

    - Hollywood
    Ooo
    I see.
    So the two x's are the boundaries, and the slope is x^3
    Not sure where the y=0 goes.
    But I think I get it, thanks
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