# integration

• Oct 10th 2011, 07:24 PM
satis
integration
I have the following question that I'm having issues solving

$\int\limits_0^1{(1+\sqrt{x})^8}dx$

I tried using uv substitution with u being ${(1+\sqrt{x})^8}$ and v being just dx. However, it quickly devolved into uv substitution after uv substition that spiralled into a massively negative number, so I must be doing something wrong. My plan was just to uv until I got that exponent down to something I could easily multiply out.

Any tips, tricks, or something I may have missed?
• Oct 10th 2011, 07:29 PM
TheChaz
Re: integration
Have you considered using the binomial theorem/expansion? It wouldn't be too ugly ...
• Oct 10th 2011, 07:36 PM
cheme
Re: integration
Make a substitution of $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$

Then, make a substitution of $s=u+1$ and $ds=du$

Doing this right will give you the following:

$2\int (s-1)s^{8}ds$

Do the required integration, then substitute back in to get the final answer.
• Oct 10th 2011, 07:47 PM
Prove It
Re: integration
Quote:

Originally Posted by cheme
Make a substitution of $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$

Then, make a substitution of $s=u+1$ and $ds=du$

Doing this right will give you the following:

$2\int (s-1)s^{8}ds$

Do the required integration, then substitute back in to get the final answer.

Or even just make the substitution $\displaystyle u = 1 + \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx$...
• Oct 11th 2011, 09:00 AM
satis
Re: integration
Thanks to you all for being of assistance. @theChaz, I was trying to avoid actually expanding that, though I suppose it's always an option.

@cheme and Prove It, looks like you two are advocating the same method, with Prove It's being a bit quicker and to the point. I'm afraid I'm not really following, though. I was wondering if you could clarify.

I understand taking the $1+\sqrt{x}$ and setting is as u. Then $du = \frac{1}{2}x^{-\frac{1}{2}}dx$.

From my understanding, this basically gets us down to $\int(u)^8dx$, so I need to use the du definition above to replace the dx. But I get stuck here. Can someone show me the rest of the substitution part?

If I try to work through it, I end up with:
$du = \frac{1}{2}x^{-\frac{1}{2}}dx$
$2du = x^{-\frac{1}{2}}dx$
I can't just move the $x^{-\frac{1}{2}}$ to the left side so I get a straight definition for dx.
• Oct 11th 2011, 09:23 AM
Siron
Re: integration
Let $1+\sqrt{x}=u \Rightarrow \frac{dx}{2\sqrt{x}}=du \Leftrightarrow dx=2\sqrt{x}du \Leftrightarrow dx=2(u-1)du$
Therefore you get:
$2\int u^{8}\cdot(u-1)du=2\int u^9du -2 \int u^8du=...$

Can you finish?

EDIT:
Now I notice you're dealing with a definite integral therefore change the integration limits or do the back-substitution at the end.
• Oct 11th 2011, 10:07 AM
satis
Re: integration
Siron, thanks for the assistance. I'm probably missing something blindingly obvious, but how do you get from $dx = 2\sqrt{x}du$ to $dx = 2(u-1)du$?

Once you substitute everything back into the integral, I follow the rest of it.

$2\int\limits_1^2{u^8(u-1)du}$
$=2\int\limits_1^2{u^9 - u^8du}$
$=2(\int\limits_1^2{u^9du} - \int\limits_1^2{u^8du})$
$=2([\frac{1}{10}u^10]\limits_1^2 - [\frac{1}{9}u^9]\limits_1^2])$
$=2([\frac{1024}{10} - \frac{1}{10}] - [\frac{512}{9} - \frac{1}{9}])$
$=\frac{1023}{5} - \frac{1022}{9}$
...jeez, what a mess. At any rate, assuming I didn't make a mistake somewhere along the line, you end up with a really ugly fraction.
• Oct 11th 2011, 10:29 AM
Siron
Re: integration
Quote:

Originally Posted by satis
Siron, thanks for the assistance. I'm probably missing something blindingly obvious, but how do you get from $dx = 2\sqrt{x}du$ to $dx = 2(u-1)du$?

Because the subtitution was $1+\sqrt{x}=u \Leftrightarrow \sqrt{x}=u-1$ (1) and you have $dx=2\sqrt{x}du$ (2), substituting (1) in (2) gives $dx=2(u-1)du$
• Oct 11th 2011, 10:36 AM
satis
Re: integration
Ah, I knew I missed something obvious. Thank you for pointing it out to me.