# Thread: improper integral

1. ## 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!

2. a) $
\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: $x>0
$
then to this integral con be improper: $
2 - p \leqslant 0 \vee 1 - p \leqslant 0 \Rightarrow \therefore p \geqslant 1
$

b) Maybe could be a better idea, find Y

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

Integrating whit respecte X, we have: $
\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: $y(x)=Ae^{\sqrt{x}}$

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