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Math Help - 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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    135
    a) <br />
\int_1^2 {\frac{1}<br />
{{x - p}}dx}  = \ln \left( {2 - p} \right) - \ln \left( {1 - p} \right)<br />

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

    b) Maybe could be a better idea, find Y

    <br />
y'(x) = \frac{{y(x)}}<br />
{{2x}} \Leftrightarrow \frac{{y'(x)}}<br />
{{y(x)}} = \frac{1}<br />
{{2x}}<br />

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

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

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