Really Hard Reduction Question (1 Viewer)

cutemouse · May 13, 2009

Define

$I_n=\int_0^\frac{a}{b} x^n \sqrt{a-bx} dx$

Where $n\ge0$ and a,b>0

Find a forumla for I_n in closed form in terms of a, b and n.

'Closed form' -- using powers and indices and factorial notation.

Thanks guys =D

Drongoski · May 13, 2009

Affinity · May 13, 2009

jm01 said:
Define

$I_n=\int_0^\frac{a}{b} x^n \sqrt{a-bx} dx$

Where $n\ge0$ and a,b>0

Find a forumla for I_n in closed form in terms of a, b and n.

'Closed form' -- using powers and indices and factorial notation.

Thanks guys =D

you can transform it to

$\sqrt{a}\left(\frac{a}{b}\right)^{n+1} \int_0^1 x^n (1-x)^{1/2}\, dx$
Then you look up the beta function on wikipedia, then the gamma function

comes out as something ugly like

$\sqrt{a}\left(\frac{a}{b}\right)^{n+1} \frac{3n!(n+1)!2^{2n+1}}{(2n+3)!}$

cutemouse · May 13, 2009

Uhh, Affinity.. thanks for your help, but how did you get your final result?

Trebla · May 13, 2009

$I_n=\int_0^\frac{a}{b} x^n \sqrt{a-bx}\,dx\\\\u=x^n\Rightarrow du=nx^{n-1}\\dv=(a-bx)^{\frac{1}{2}}\,dx\Rightarrow v=-\dfrac{2(a-bx)^{\frac{3}{2}}}{3b}\\\\I_n=\dfrac{2n}{3b}\int_0^\frac{a}{b}(a-bx)^{\frac{3}{2}}x^{n-1}\,dx\\\\= \dfrac{2n}{3b}\int_0^\frac{a}{b} \sqrt{a-bx}(a-bx)x^{n-1}\,dx\\\\= \dfrac{2n}{3b}\int_0^\frac{a}{b}(ax^{n-1}\sqrt{a-bx}-bx^n\sqrt{a-bx})\,dx\\\\=\dfrac{2an}{3b}I_{n-1}-\dfrac{2n}{3}I_n\\\\\therefore (1+\dfrac{2n}{3})I_n=\dfrac{2an}{3b}I_{n-1}\\\\\Rightarrow I_n=\dfrac{2an}{b(2n+3)}I_{n-1}$

$I_n=\dfrac{2an}{b(2n+3)}I_{n-1}\\\\I_{n-1}=\dfrac{2a(n-1)}{b(2n+1)}I_{n-2}\\\\I_{n-2}=\dfrac{2a(n-2)}{b(2n-1)}I_{n-3}\\\\.\\.\\.\\.\\I_1=\frac{2a(1)}{b(2+3)}I_0\\\\$But $I_0=\displaystyle\int_0^\frac{a}{b} \sqrt{a-bx}\,dx\\\\= \left [\dfrac{-2(a-bx)^{\frac{3}{2}}}{3b} \right ]_0^\frac{a}{b}\\\\= \dfrac{2a^{\frac{3}{2}}}{3b}\\\\\therefore I_1=\frac{2a(1)}{b(2+3)}\dfrac{2a^{\frac{3}{2}}}{3b}\\\\$Multiplying all equations gives$\\\\I_n = \dfrac{(2a)^{n+1}n!}{b^{n+1}(2n+3)(2n+1)(2n-1).....7.5.3.1}a^{\frac{1}{2}}\\\\=\dfrac{a^{\frac{1}{2}}2^{n+1}a^{n+1}n!}{b^{n+1}}\dfrac{1}{(2n+3)(2n+1)(2n-1).....7.5.3.1}\\\\ $Multiplying top and bottom by $(2n+2)(2n)(2n-2)....4.2=2^{n+1}(n+1)!\\\\I_n = \dfrac{a^{\frac{1}{2}}2^{2n+2}a^{n+1}n!(n+1)!}{b^{n+1}(2n+3)!}$

Not sure if this is corect. Check my working...

2000th post!!!!

Really Hard Reduction Question (1 Viewer)

cutemouse

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Drongoski

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Affinity

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cutemouse

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Trebla

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