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Math Help - Integration- Area under curves

  1. #1
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    Integration- Area under curves

    I think this question is pretty basic, but I would be very happy if you could help

    Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1

    Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

    Thanks!!
    Lachlan
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by auonline View Post
    I think this question is pretty basic, but I would be very happy if you could help

    Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1

    Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

    Thanks!!
    Lachlan
    Yes the required area is:

    A=\int_{x=-1}^1 \frac{1}{(x-2)^2}+1 \; dx

    what method you use to evaluate it is up to you and/or your teacher.

    CB
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  3. #3
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    Ok thanks
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by auonline View Post

    Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1
    Another way:

    \lim_{n\to\infty}\sum_{i=1}^n\left[\frac{1}{[(-1+\frac{2i}{n})-2]^2}+1\right]\left(\frac{2}{n}\right)
    Last edited by VonNemo19; July 13th 2009 at 01:12 PM.
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    Another way:

    \lim_{n\to\infty}\sum_{i=1}^n\left[\frac{1}{[(-1+\frac{2i}{n})-2]^2}+1\right]\left(\frac{2}{n}\right)
    Ah, the easy way!
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  6. #6
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    Quote Originally Posted by auonline View Post
    I think this question is pretty basic, but I would be very happy if you could help

    Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1

    Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

    Thanks!!
    Lachlan
    If you let u= x-2, differentiating both sides give du= dx (was that what you meant by "finds the derivate"); when x= -1 , u= -1-2= -3, and when x= 1, u= 1-2= -1. Now replace each of those things in your integral by its value in terms of u, \int_{-1}^1 ((x-2)^2 + 1)dx becomes \int_{-3}^{-1} (u^{-2}+ 1) du. That is NOT some "random hard to understand process". That is a very logical, standard method known as "integration by substitution".
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Ah, the easy way!
    I think so. I realize that this is not as quick as u-substitution, but it is good to see things from all sides when trying to tackle a problem.
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