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Thread: improper integral

  1. #1
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    improper integral

    hey everyone, got 2 questions.

    the first one asks for determining all values of p for which dx/(x-p) is improper evaluated from 1 to 2.

    another question asks about finding an equation for the integral curve that passes through the point (2,1) for the differential equation y'=y/2x and how do you go about graphing y'=y/2x?

    cheers!
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  2. #2
    Member Nacho's Avatar
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    a) $\displaystyle
    \int_1^2 {\frac{1}
    {{x - p}}dx} = \ln \left( {2 - p} \right) - \ln \left( {1 - p} \right)
    $

    we know that the domain the logarythm function is to: $\displaystyle x>0
    $ then to this integral con be improper: $\displaystyle
    2 - p \leqslant 0 \vee 1 - p \leqslant 0 \Rightarrow \therefore p \geqslant 1
    $

    b) Maybe could be a better idea, find Y

    $\displaystyle
    y'(x) = \frac{{y(x)}}
    {{2x}} \Leftrightarrow \frac{{y'(x)}}
    {{y(x)}} = \frac{1}
    {{2x}}
    $

    Integrating whit respecte X, we have: $\displaystyle
    \int {\frac{{y'(x)}}
    {{y(x)}}dx} = \frac{1}
    {2}\int {\frac{1}
    {x}dx} \Leftrightarrow \ln \left( {y(x)} \right) = \frac{1}
    {2}\ln x + C
    $

    Hence: $\displaystyle y(x)=Ae^{\sqrt{x}}$

    But curve passes through the point (2,1) then $\displaystyle
    y(2) = 1 \Leftrightarrow 1 = Ae^{\sqrt 2 }
    $ $\displaystyle
    \therefore y(x) = \exp \left( {x - \sqrt 2 } \right)
    $
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