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Math Help - area under the curve

  1. #1
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    area under the curve

    Find the area of region under the curve  y = \frac{1}{x^2+x} to the right of  x = 1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    Find the area of region under the curve  y = \frac{1}{x^2+x} to the right of  x = 1
    You are looking for
    \int_1^{\infty} \frac{dx}{x^2 + x}

    Hint: x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}

    Put this in the denominator and use the substitution y = x + \frac{1}{2}. Does this integral look more familiar?

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    You are looking for
    \int_1^{\infty} \frac{dx}{x^2 + x}

    Hint: x^2 + x = x^2 + x + \frac{1}{4} - \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2 - \frac{1}{4}

    Put this in the denominator and use the substitution y = x + \frac{1}{2}. Does this integral look more familiar?

    -Dan
    Notice that 1/(x^2+x)=1/x-1/(x+1)
    Attached Thumbnails Attached Thumbnails area under the curve-ccc.gif  
    Last edited by curvature; June 20th 2007 at 11:47 AM.
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  4. #4
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    Quote Originally Posted by curvature View Post
    Notice that 1/(x^2+x)=1/x-1/(x+1)
    the region looks like this
    Attached Thumbnails Attached Thumbnails area under the curve-xx.gif  
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by curvature View Post
    To viet: finally, you take the limit with \infty, then it's done.
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  6. #6
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    i got the final answer to be  ln 2 . thanks y'all
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