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Help with Integration algebra!

  1. Aug 8, 2005 #1
    Howdy felles.

    I have a question which I find rather confusing, and that is to prove that the area between two curves is given by [tex]A=\frac{n-1}{n+1} [/tex] where n is the power of X.


    The equation is [tex]A=\int X^\frac{1}{n} dx - \int X^n dx[/tex]

    Could someone plz show me how to get this to equal to [tex]A=\frac{n-1}{n+1} [/tex]

    I had a very diffrent answer. Plz help

    Cheers
     
  2. jcsd
  3. Aug 8, 2005 #2
    Are there limits on those integrals ?
     
  4. Aug 9, 2005 #3
    The limits are from 0 to 1
     
  5. Aug 9, 2005 #4

    lurflurf

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    Homework Helper

    First do the integrals
    [tex]=\int_0^1 x^\frac{1}{n} dx[/tex]
    and
    [tex]=\int_0^1 x^n dx[/tex]
    Then simplify and get a common denominator.
    What did you get for the integrals?
     
    Last edited: Aug 9, 2005
  6. Aug 9, 2005 #5
    I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one plz check where I made mistake>

    [tex]A=\int X^\frac {1}{n} dx - \int X^n dx[/tex]
    [tex]A=\frac {[X^\frac {1}{n}+1]}{\frac {1}{n}+1} - \frac {[X^n+1}{n+1]}[/tex] Cant seem to make any progress from here onwards. All I do is make a massive mess of it and I have to do it again without any good signs.

    Would really appriciate if anyone could help me.
     
  7. Aug 9, 2005 #6
    I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one plz check where I made mistake>

    [tex]A=\int X^\frac {1}{n} dx - \int X^n dx[/tex]
    [tex]A=\frac {X^\frac {1}{n+1}}{\frac {1}{n}+1} - \frac {X^n+1}{n+1}[/tex] Cant seem to make any progress from here onwards. All I do is make a massive mess of it and I have to do it again without any good signs.

    Would really appriciate if anyone could help me.
     
  8. Aug 9, 2005 #7
    Also the limits of all these integrations are 0 to 1
     
  9. Aug 9, 2005 #8

    VietDao29

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    Homework Helper

    Okay, you should remember that:
    [tex]\int^{\beta}_{\alpha} f(x)dx = F(\beta) - F(\alpha)[/tex]
    So I'll do for you the first one. And you can follow it to do the second:
    [tex]\int^{1}_{0} x ^ {\frac{1}{n}}dx = \left[ \frac{x ^ {\frac{1 + n}{n}}}{\frac{1 + n}{n}} \right] ^{1}_{0} = \frac{1 ^ {\frac{n + 1}{n}} - 0 ^ {\frac{n + 1}{n}}}{\frac{1 + n}{n}} = \frac{n}{1 + n}[/tex]
    So can you do the second integral?
    Viet Dao,
     
  10. Aug 9, 2005 #9
    I am still having abit trouble .

    I am sposed to have [tex]A=\frac {n}{1+n} - \int^{o}_{1} X^n dx [/tex]

    which I ended up getting [tex]\int^{1}_{0} X^n dx = \frac {1^n^+^1 - 0^n^+^1}{n+1}[/tex]
    does it [tex]=\frac {n}{n+1}[/tex]
     
  11. Aug 9, 2005 #10

    VietDao29

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    Homework Helper

    Nope. Your integral is 'nearly' correct. The last part is wrong.
    Remember : [tex]1 ^ m = 1, m \in \mathbb{R}[/tex]
    [tex]0 ^ m = 0, m \in \mathbb{R} - \{0\}[/tex]
    So [tex]1 ^ {n + 1} = 1[/tex]
    Viet Dao,
     
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