integral with square root

**chocolatelover** · September 2nd 2007, 12:16 PM

Hi everyone,

In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

Int. 0-1 r^3/sq. 4+r^2 dr

t=r^3
tu=3r^2dr
r^2dr=tu/3

Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

Thank you

**Krizalid** · September 2nd 2007, 12:31 PM

Set $u=\sqrt{4+r^2}$

**red_dog** · September 2nd 2007, 12:32 PM

You better use the substitution
$\sqrt{4+r^2}=u\Rightarrow 4+r^2=u^2\Rightarrow rdr=udu$
$r=0\Rightarrow u=2$
$r=1\Rightarrow u=\sqrt{5}$
Now
$\displaystyle\int_0^1r^2\sqrt{4+r^2}\cdot rdr=\int_2^{\sqrt{5}}(u^2-4)u^2du$
and you can continue.

**chocolatelover** · September 3rd 2007, 12:44 PM

Hi,

Can you explain how you got "

" please?

Also, your final answer would be -1 right?

This is what I did:

u squared -4=(2 squared-4)-(sq. 5) sq. -4=-1

Thank you

**Krizalid** · September 3rd 2007, 01:11 PM

$u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr$ , so $u\,du=r\,dr$

Does that make sense?

**chocolatelover** · September 3rd 2007, 01:15 PM

Yes, thank you very much

**mehranhydary** · March 19th 2010, 12:16 PM

Originally Posted by Krizalid

$u=\sqrt{4+r^2}\implies u^2=4+r^2\,\therefore\,2u\,du=2r\,dr$ , so $u\,du=r\,dr$

Does that make sense?

how about root (4-2t)

**mehranhydary** · March 19th 2010, 12:28 PM

Originally Posted by chocolatelover

Hi everyone,

In order to solve this integral I would need to do a u substitution and then use integration of parts, right?

Int. 0-1 r^3/sq. 4+r^2 dr

t=r^3
tu=3r^2dr
r^2dr=tu/3

Then I would finish the u substitution and then do integration of parts, right? Or could I just do a u substitution?

Thank you

sssss

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