# integral with square root

• Sep 2nd 2007, 01:16 PM
chocolatelover
integral with square root
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
• Sep 2nd 2007, 01:31 PM
Krizalid
Set $u=\sqrt{4+r^2}$
• Sep 2nd 2007, 01:32 PM
red_dog
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.
• Sep 3rd 2007, 01:44 PM
chocolatelover
Hi,

Can you explain how you got "http://www.mathhelpforum.com/math-he...6b4910ec-1.gif" please?

This is what I did:

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

Thank you
• Sep 3rd 2007, 02:11 PM
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?
• Sep 3rd 2007, 02:15 PM
chocolatelover
Yes, thank you very much
• Mar 19th 2010, 01:16 PM
mehranhydary
Quote:

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?

• Mar 19th 2010, 01:28 PM
mehranhydary
Quote:

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

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