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Math Help - Quick Integration

  1. #1
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    Exclamation 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
    Last edited by nath_quam; June 23rd 2006 at 05:55 AM.
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  2. #2
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    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
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  3. #3
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    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:

    <br />
\int_{-1}^0 x^n\sqrt{1+x}\ dx<br />

    surly?

    RonL
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  4. #4
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    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.
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  5. #5
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    i was asking bout what captain wrote i really shoud learn how to make it look good
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  6. #6
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    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
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  7. #7
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    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 !
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  8. #8
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    Thanks that would be good if u could perfect cos i get stuck when i do it
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  9. #9
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    Quote Originally Posted by CaptainBlack
    He is asking about:

    <br />
\int_{-1}^0 x^n\sqrt{1+x}\ dx<br />

    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

    <br />
\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<br />
.

    Putting the limits back in:

    <br />
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<br />
.

    Simplifying:

    <br />
I_n=\int_{-1}^0 x^n\ \sqrt{1+x} \ dx= <br />
 - \int_{-1}^0 \frac{2}{3}(1+x)\sqrt{1+x}\ n\ x^{n-1} dx<br />
    <br />
= - \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]<br />
    <br />
=- \frac{2}{3}n [I_{n-1}+I_n]<br />
    .

    So the recurrence relation for I_n is:

    <br />
I_n=-\frac{\frac{2}{3}n}{1-\frac{2}{3}n}I_{n-1}<br />

    RonL

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