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 u^{10}/10 interval 1 to 0

-1 (0^{10}/10 - 1^{10}/10)

=1/10

Am I correct?

Printable View

- Mar 3rd 2013, 09:44 PMSteelers72Definite 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 u^{10}/10 interval 1 to 0

-1 (0^{10}/10 - 1^{10}/10)

=1/10

Am I correct? - Mar 3rd 2013, 10:50 PMsbhatnagarRe: Definite integral question
$\displaystyle \int_0^1 (1-x)^9 dx = \frac{\Gamma(1)\Gamma(10)}{\Gamma(11)}=\frac{9!}{1 0!}=\frac{1}{10}$

Yes, your answer is correct! - Mar 3rd 2013, 10:52 PMearbothRe: Definite integral question
You certainely noticed that $\displaystyle \frac1{(1-x)^9}$ is not defined for x = 1, that means the graph of $\displaystyle f(x)=\frac1{(1-x)^9}$ has a vertical asymptote x = 1. Therefore the definite integral doesn't exist.

EDIT: Sorry, I misunderstood your question completely!