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Math Help - How would you tackle this question? [Area]

  1. #1
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    Question How would you tackle this question? [Area]

    I'm not asking you to do it for me (unless you want to ), but i just dont know what to do, so what would you do..?

    Heres the Q:

    For the function , the ratio of area A:Area B is n:1, where A is the area from y = 0 to 1 and the y axis, and B is the area from x = 0 to 1 and the x axis. As shown:


    Is this ratio true for the general case from x = a to b such that a < b, and for the regions defined below?

    Area A: , , and the y axis
    Area B: , , and the x axis

    If so prove it; if not why not?

    ___

    Once again, I'm not asking you to do it for me, I just need help as to what to do

    I've proven that the ratio holds for like x = 0 to 5, or 2 to 7, but what else do i need to do?/ What is the question really asking for
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  2. #2
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    Hello, jhomie!

    For the function: y = x^n, the ratio of area A to area B is n:1
    where A is the area from y = 0\text{ to }1 and the y-axis,
    and B is the area from x = 0\text{ to }1 and the x-axis.
    Code:
          |
         1+ - - - - -*
          |::::::::::|
          |::: A :::*|
          |::::::::* |
          |::::::*   |
          |:::*   B  |
        --*----------+--
          |           1

    Area A

    A \;=\;\int^1_0\left(1 - x^n\right)\,dx \;=\;x - \frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;1 - \frac{1}{n+1} \;=\;\frac{n}{n+1}


    Area B

    B \;=\;\int^1_0 x^n\,dx \;=\;\frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;\frac{1}{n+1}


    Therefore: . \frac{A}{B} \;=\;\frac{\frac{n}{n+1}}{\frac{1}{n+1}} \;=\;\frac{n}{1}


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  3. #3
    Newbie
    Joined
    Nov 2008
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    Quote Originally Posted by Soroban View Post
    Hello, jhomie!


    Area A

    A \;=\;\int^1_0\left(1 - x^n\right)\,dx \;=\;x - \frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;1 - \frac{1}{n+1} \;=\;\frac{n}{n+1}


    Area B

    B \;=\;\int^1_0 x^n\,dx \;=\;\frac{x^{n+1}}{n+1}\,\bigg]^1_0 \;=\;\frac{1}{n+1}


    Therefore: . \frac{A}{B} \;=\;\frac{\frac{n}{n+1}}{\frac{1}{n+1}} \;=\;\frac{n}{1}



    yea, i understand that. \frac{n}{1} = n:1 right?

    But that is just for 0 to 1, theres still a and b both -ve, a -ve b positive, then changing n so that it is positive integer, negative integers, fraction and irrational. Is that what i'm supposed to do?

    For y=x^\frac{4}{3} i cant seem to get the same ratio. I might be doing something wrong, but dunno
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