The problem asks to derive the following formula $\sum^n_{k=1} f(k) = [n]f(n)-\int^n_1 f'(x) [x] dx$. Assuming $f'$ is continuous, where $[x]$ is the greatest integer in $x$
I know to use the integration by parts formula for RS integral, and if $\alpha$ is a step function, the integral reduces to $\sum^n_1 f(x_i)(\alpha(x_i^+)-\alpha(x_i^-))$, but I always have one more term $f(1)[1]$. It seems that the lower bound could not be gone.
Thanks in advanced!
The integrals in the problem appear to be written just as Riemann integrals (no $\alpha$).
We have
$$\int^n_1 f'(x) [x]\, dx=\int_1^2 f'(x)\,dx+2\int_2^3 f'(x)\,dx + \cdots + (n-1)\int_{n-1}^n f'(x)\,dx$$ which in turn equals $$f(2)-f(1)+2(f(3)-f(2))+3(f(4)-f(3))+\cdots+(n-1)(f(n)-f(n-1))$$ If we look at how the terms cancel each other, we have $$-\sum_{k=1}^{n-1} f(k)+(n-1)f(n)$$ so by subtracting from $nf(n)$ we get the desired sum.