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 $\displaystyle \int_0^2 f'(x) ~dx$
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.