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- Mar 12th 2009, 03:58 AMchamprockIntegration of greatest integer function
- Mar 12th 2009, 04:05 AMMoo
Hello,

Separate the integral :

$\displaystyle \int_0^2 [x]^n f'(x) ~dx=\int_0^1 [x]^n f'(x) ~dx+\int_1^2 [x]^n f'(x)~dx$

if x is between 0 and 1, then [x]=0

if x is between 1 and 2, then [x]=1

so your integral is actually $\displaystyle \int_1^2 f'(x) ~dx$, if $\displaystyle n>0$

if n=0, then the integral is $\displaystyle \int_0^2 f'(x) ~dx$ - Mar 12th 2009, 04:16 AMchamprock
but its actually [X]^N whcih is most confusing. (mod raised to Nth power).

so, it cant be seperated that easily. - Mar 12th 2009, 04:18 AMMoo
- Mar 12th 2009, 04:29 AMchamprock
hmm.. lets just take an example. assume that n=5 and x=1.5, then 1.5^5 = 7.59..

- Mar 12th 2009, 07:37 AMKrizalid
Let $\displaystyle k\ge2,$ be an integer then $\displaystyle \int_{0}^{k}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=\sum\limits_{m=0}^{k-1}{\int_{m}^{m+1}{m^{n}f'(x)}\,dx}=\sum\limits_{m= 1}^{k-1}{m^{n}\big(f(m+1)-f(m)\big)}.$

Hence, for $\displaystyle k=2$ it's $\displaystyle \int_{0}^{2}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=f(2)-f(1),$ which is the value of the integral. - Mar 12th 2009, 09:07 AMMoo