convergent and divergent improper integrals

**saiyanmx89** · March 8th 2009, 08:41 AM

Did I do this one right? :

$\int \frac{4}{\sqrt{6-x}}dx$ [0,6]

$= lim \int \frac{4}{\sqrt{6-x}}dx$ [0,6]
$x->6^-$

$= lim (\frac{4}{(6-x)^2})| [0,b]$
$x->6^-$

$= infinity - \frac{1}{9} = diverges?$

**skeeter** · March 8th 2009, 09:15 AM

Originally Posted by saiyanmx89

Did I do this one right? :

$\int \frac{4}{\sqrt{6-x}}dx$ [0,6]

$= lim \int \frac{4}{\sqrt{6-x}}dx$ [0,6]
$x->6^-$

$= lim (\frac{4}{(6-x)^2})| [0,b]$
$x->6^-$

$= infinity - \frac{1}{9} = diverges?$

your antiderivative is incorrect ...

$\int \frac{4}{\sqrt{6-x}} \, dx = -8\sqrt{6-x} + C$

Thread: convergent and divergent improper integrals

LinkBack

Thread Tools

Display

convergent and divergent improper integrals

Similar Math Help Forum Discussions

Improper integrals of type I: convergent or divergent?

Improper integral (convergent/divergent)

Improper integral convergent or divergent?

Improper integral. Convergent or divergent?

Improper integral, tarctan(t)/(1+t^2)^2, from 0 to infinity, convergent or divergent?

Search Tags