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Math Help - Finding the total area between the function and the x-axis

  1. #1
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    Finding the total area between the function and the x-axis

    I'm supposed to find the area in the region between the function and the x-axis....

    y = x^(1/3), [-1,8]

    I divided it up into two regions (-1 to 0 -> A1 and 0 to 8 -> A2). I found A2 to be 12 and A1 to be |-3/4|. My teacher worked the problem on the review sheet and she got |-1| for A1. How is this possible?!

    We both have the same equation:

    A1 = (3/4)x^(4/3) evaluated from 0 to -1. Obviously x=0 will yield 0, and when I plug in -1 I get -3/4! Am I just completely missing something?

    For the total area, I get 12.75 units, and my teacher got 13.
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  2. #2
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    Quote Originally Posted by tom ato View Post
    I'm supposed to find the area in the region between the function and the x-axis....

    y = x^(1/3), [-1,8]

    I divided it up into two regions (-1 to 0 -> A1 and 0 to 8 -> A2). I found A2 to be 12 and A1 to be |-3/4|. My teacher worked the problem on the review sheet and she got |-1| for A1. How is this possible?!

    We both have the same equation:

    A1 = (3/4)x^(4/3) evaluated from 0 to -1. Obviously x=0 will yield 0, and when I plug in -1 I get -3/4! Am I just completely missing something?

    For the total area, I get 12.75 units, and my teacher got 13.
    I'm with you. You'll notice that your function is symmetry about the origin so A1 = \int_0^1 x^{1/3}\,dx = \left. \frac{3}{4} x^{4/3} \right |_0^1 = 3/4 as you say. Further

    A2 = \int_0^8 x^{1/3}\,dx = \left. \frac{3}{4} x^{4/3} \right |_0^8= 12
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  3. #3
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    A_1 = \int_{-1}^0 (0-x^{1/3})dx=\left[-\frac{3}{4}x^{4/3}\right]_{-1}^0=\frac{3}{4}
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  4. #4
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    Quote Originally Posted by tom ato View Post
    My teacher worked the problem on the review sheet and she got |-1| for A1.
    she forgot the factor 1/(n+1) perhaps.
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