integral of the function $[x]$ stands for integral part of $x$, $n$ is positive and $f'$ is the derivative of $f$
$$\int_{0}^{2} [x]^n.f'(x)dx$$
I have done it by separating the limits
$$ =>\int_{0}^{1} [x]^n.f'(x)dx_\ + _\ \int_{1}^{2} [x]^n.f'(x)dx $$
$$=>(2)^n f(2)  -f(1)$$
but the answer given is $f(2)+f(1)$

$[x]=1$ for $1 <x<2$ (assuming that $[x]$ is the floor function). So the correct value is $f(2)-f(1)$. Your answer and the given answer are both wrong!
