# integration of binomial expansion

• Dec 10th 2008, 09:03 AM
hmmmm
integration of binomial expansion
how would you integrate the general binomial expansion??
http://www.mathhelpforum.com/math-he...2c215de0-1.gif
thanks for and help
• Dec 10th 2008, 11:33 AM
fobos3
$\displaystyle \int (a\times x+b)^n \, dx=\frac{(a x+b)^{n+1}}{a(n+1)}$

If $\displaystyle F (x) = \int f(x)dx$ then:
F(ax+b)=F(x)/a
• Dec 10th 2008, 02:15 PM
hmmmm
thanks
thanks but i was more asking how you would integrate the right hand side of the equation? sorry
• Dec 10th 2008, 02:39 PM
Mathstud28
Quote:

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 $\displaystyle \int\sum_{k=0}^{n}{n\choose{k}}x^{n-k}y^k=\sum_{k=0}^{n}\int\left\{{n\choose{k}}x^{n-k}y^k\right\}$
• Dec 11th 2008, 12:46 PM
hmmmm
thanks
so how do we integrate this?
• Dec 11th 2008, 01:20 PM
Greengoblin
$\displaystyle \sum_{k=0}^{n} {n\choose{k}}x^{n-k}y^k$

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

So taking:

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

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