# Math Help - Definite integral question

1. ## Definite integral question

Integral from 0 to 1 (1-x)9

I did u substitution
u=1-x
du=-1dx
du/-1 =dx

get new intervals
1-1=0
1-0=1

-1 integral u^9 du
-1 u10/10 interval 1 to 0

-1 (010/10 - 110/10)

=1/10

Am I correct?

2. ## Re: Definite integral question

$\int_0^1 (1-x)^9 dx = \frac{\Gamma(1)\Gamma(10)}{\Gamma(11)}=\frac{9!}{1 0!}=\frac{1}{10}$

3. ## Re: Definite integral question

Originally Posted by Steelers72
Integral from 0 to 1 (1-x)9

I did u substitution
u=1-x
du=-1dx
du/-1 =dx

get new intervals
1-1=0
1-0=1

-1 integral 1/u^9 du <-- unfortunately wrong
-1 u10/10 interval 1 to 0

-1 (010/10 - 110/10)

=1/10

Am I correct?
You certainely noticed that $\frac1{(1-x)^9}$ is not defined for x = 1, that means the graph of $f(x)=\frac1{(1-x)^9}$ has a vertical asymptote x = 1. Therefore the definite integral doesn't exist.

EDIT: Sorry, I misunderstood your question completely!