Quick Integration

Printable View

June 23rd 2006, 02:54 AM
nath_quam

Quick Integration

How do i find and what is the recurrence formula for (x^n)(1+x)^1/2.dx between the x terminals 0 and -1 ? Thanks Nath
June 23rd 2006, 09:41 AM
ThePerfectHacker

Quote:

Originally Posted by nath_quam

How do i find and what is the recurrence formula for (x^n)(1+x)^1/2.dx between the x terminals 0 and -1 ? Thanks Nath

First you problably mean that,
$0\leq x\leq 1$ otherwise the function is not defined, $x=-1$ which means that ,
$\int_{-1}^0 \frac{x^n}{\sqrt{1+x}}dx$ makes no sense at all!.

Assuming you mean,
$\int_0^1\frac{x^n}{\sqrt{1+x}}dx$
Use integration by parts,
let, $u=x^n$ then, $u'=nx^{n-1}$ and, $v'=\frac{1}{\sqrt{1+x}}$ then, $v=2\sqrt{1+x}$
Thus,
$\left 2x^n\sqrt{1+x} \right|^1_0-\int_0^1 2nx^{n-1}\sqrt{1+x}dx$
Thus,
$2\sqrt{2}-2n\int_0^1x^{n-1}\sqrt{1+x}dx$
The trick is to express,
$\sqrt{1+x}$ as, $\frac{1+x}{\sqrt{1+x}}$
Thus,
$2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$
Working with the integral we have,
$\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx=\int_0^1\frac{x^{n-1}+x^n}{\sqrt{1+x}}$ = $\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\int^1_0\frac{x^n}{\sqrt{1+x}}dx$
Let,
$\chi=\int_0^1\frac{x^n}{\sqrt{1+x}}$
Then, as we have shown,
$\chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$
Which is,
$\chi=2\sqrt{2}-2n\left(\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\chi\right)$
Thus,
$\chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx-2n\chi$
Thus,
$\chi(2n+1)=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$
Thus,
$\chi=\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$
Thus, the recurrence formula is,
$\boxed{ \int_0^1\frac{x^n}{\sqrt{1+x}}dx =\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx}$

---
Credits:
I learend the box trick from Soroban
June 23rd 2006, 09:59 AM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

First you problably mean that,
$0\leq x\leq 1$ otherwise the function is not defined, $x=-1$ which means that ,
$\int_{-1}^0 \frac{x^n}{\sqrt{1+x}}dx$ makes no sense at all!.

He is asking about:

$ \int_{-1}^0 x^n\sqrt{1+x}\ dx $

surly?

RonL
June 23rd 2006, 10:02 AM
ThePerfectHacker

Sorry, I see things differently sometimes.
My solution to the above problem follows the same idea.
Almost identical.
Try to find the solution yourself, if not I can finish it.
June 23rd 2006, 05:25 PM
nath_quam

i was asking bout what captain wrote i really shoud learn how to make it look good
June 23rd 2006, 05:34 PM
malaygoel

Quote:

Originally Posted by nath_quam

i was asking bout what captain wrote i really shoud learn how to make it look good

Substitute (1+x)=t
June 24th 2006, 06:52 PM
ThePerfectHacker

nath quam
Did you get the recrussive relationship yet?
I can integrate it for you if you like.
After all, I am the quickest integrator in the west !
June 24th 2006, 10:44 PM
nath_quam

Thanks that would be good if u could perfect cos i get stuck when i do it
June 24th 2006, 11:31 PM
CaptainBlack

Quote:

Originally Posted by CaptainBlack

He is asking about:

$ \int_{-1}^0 x^n\sqrt{1+x}\ dx $

surly?

RonL

Let:

$I_n=\int_{-1}^0 x^n\sqrt{1+x}\ dx$ .

Now we are going to use integration by parts, which has the schema:

$\int u\ dv=uv-\int v\ du$ .

Here we let: $dv=\sqrt{1+x}$ , and $u=x^n$ , so:

$v=\frac{2}{3}(1+x)^{3/2}=\frac{2}{3}(1+x)\sqrt{1+x}$ ,

and:

$du=n\ x^{n-1}$ .

So

$ \int x^n\ \sqrt{1+x} \ dx=x^n\ \frac{2}{3}(1+x)^{3/2}$ $- \int \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx $ .

Putting the limits back in:

$ I_n=\int_{-1}^0 x^n\ \sqrt{1+x} \ dx=\left[x^n\ \frac{2}{3}(1+x)^{3/2}\right]_{-1}^0$ $- \int_{-1}^0 \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx $ .

Simplifying:

$ I_n=\int_{-1}^0 x^n\ \sqrt{1+x} \ dx=$ $ - \int_{-1}^0 \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx $

$ = - \frac{2}{3}n \left[\int_{-1}^0 \sqrt{1+x}\ x^{n-1} dx+\int_{-1}^0 \sqrt{1+x}\ x^{n} dx\right] $
$ =- \frac{2}{3}n [I_{n-1}+I_n] $
.

So the recurrence relation for $I_n$ is:

$ I_n=-\frac{\frac{2}{3}n}{1-\frac{2}{3}n}I_{n-1} $

RonL

(you will need to check this carefully as there are too many steps to
guarantee there is no mistake)