integration of binomial expansion

**hmmmm** · December 10th 2008, 09:03 AM

how would you integrate the general binomial expansion??

thanks for and help

**fobos3** · December 10th 2008, 11:33 AM

$\int (a\times x+b)^n \, dx=\frac{(a x+b)^{n+1}}{a(n+1)}$

If $F (x) = \int f(x)dx$ then:
F(ax+b)=F(x)/a

**hmmmm** · December 10th 2008, 02:15 PM

thanks but i was more asking how you would integrate the right hand side of the equation? sorry

**Mathstud28** · December 10th 2008, 02:39 PM

Originally Posted by hmmmm

thanks but i was more asking how you would integrate the right hand side of the equation? sorry

Since the sum is finite you may interchange the summation and the integral so $\int\sum_{k=0}^{n}{n\choose{k}}x^{n-k}y^k=\sum_{k=0}^{n}\int\left\{{n\choose{k}}x^{n-k}y^k\right\}$

**hmmmm** · December 11th 2008, 12:46 PM

so how do we integrate this?

**Greengoblin** · December 11th 2008, 01:20 PM

$\sum_{k=0}^{n} {n\choose{k}}x^{n-k}y^k$

$= x^n+nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+\cdots +\frac{n(n-1)\cdots(n-r+1)}{r!}x{n-r}y^r+\cdots+y^n$

So taking:

$\int\sum_{k=0}^{n}{n\choose{k}}x^{n-k}y^kdx$

Is just a case of using the addition and power rules for integrals on this series.

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integration of binomial expansion

thanks

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