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

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- Jun 23rd 2006, 02:54 AMnath_quamQuick 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

- Jun 23rd 2006, 09:41 AMThePerfectHackerQuote:

Originally Posted by**nath_quam**

$\displaystyle 0\leq x\leq 1$ otherwise the function is not defined, $\displaystyle x=-1$ which means that ,

$\displaystyle \int_{-1}^0 \frac{x^n}{\sqrt{1+x}}dx$ makes no sense at all!.

Assuming you mean,

$\displaystyle \int_0^1\frac{x^n}{\sqrt{1+x}}dx$

Use integration by parts,

let, $\displaystyle u=x^n$ then, $\displaystyle u'=nx^{n-1}$ and, $\displaystyle v'=\frac{1}{\sqrt{1+x}}$ then, $\displaystyle v=2\sqrt{1+x}$

Thus,

$\displaystyle \left 2x^n\sqrt{1+x} \right|^1_0-\int_0^1 2nx^{n-1}\sqrt{1+x}dx$

Thus,

$\displaystyle 2\sqrt{2}-2n\int_0^1x^{n-1}\sqrt{1+x}dx$

The trick is to express,

$\displaystyle \sqrt{1+x}$ as, $\displaystyle \frac{1+x}{\sqrt{1+x}}$

Thus,

$\displaystyle 2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$

Working with the integral we have,

$\displaystyle \int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx=\int_0^1\frac{x^{n-1}+x^n}{\sqrt{1+x}}$=$\displaystyle \int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\int^1_0\frac{x^n}{\sqrt{1+x}}dx$

Let,

$\displaystyle \chi=\int_0^1\frac{x^n}{\sqrt{1+x}}$

Then, as we have shown,

$\displaystyle \chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}(1+x)}{\sqrt{1+x}}dx$

Which is,

$\displaystyle \chi=2\sqrt{2}-2n\left(\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx+\chi\right)$

Thus,

$\displaystyle \chi=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx-2n\chi$

Thus,

$\displaystyle \chi(2n+1)=2\sqrt{2}-2n\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$

Thus,

$\displaystyle \chi=\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx$

Thus, the recurrence formula is,

$\displaystyle \boxed{ \int_0^1\frac{x^n}{\sqrt{1+x}}dx =\frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}\int_0^1\frac{x^{n-1}}{\sqrt{1+x}}dx}$

---

Credits:

I learend the box trick from Soroban - Jun 23rd 2006, 09:59 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle

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

$

surly?

RonL - Jun 23rd 2006, 10:02 AMThePerfectHacker
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. - Jun 23rd 2006, 05:25 PMnath_quam
i was asking bout what captain wrote i really shoud learn how to make it look good

- Jun 23rd 2006, 05:34 PMmalaygoelQuote:

Originally Posted by**nath_quam**

- Jun 24th 2006, 06:52 PMThePerfectHacker
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 ! - Jun 24th 2006, 10:44 PMnath_quam
Thanks that would be good if u could perfect cos i get stuck when i do it

- Jun 24th 2006, 11:31 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

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

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

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

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

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

and:

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

So

$\displaystyle

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

$.

Putting the limits back in:

$\displaystyle

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

$.

Simplifying:

$\displaystyle

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

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

$

$\displaystyle.

= - \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]

$

$\displaystyle

=- \frac{2}{3}n [I_{n-1}+I_n]

$

So the recurrence relation for $\displaystyle I_n$ is:

$\displaystyle

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)