Integral wrt floor(x) What is the definite integral of $f(x)=x^2+1$ with respect to the differential of $\lfloor x\rfloor$ i.e ($d\lfloor x\rfloor$) from $0$ to $2$?
I tried to multiply and divide dx by then $d\lfloor x\rfloor/dx = 0$.
How do I approach it?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Integrate by parts ( Why not ? ):
\begin{align}
&\bbox[#ffd,5px]{%
\int_{0}^{2}\pars{x^{2} + 1}\dd\lfloor x\rfloor}
\\[3mm] = &\
\overbrace{\left.\left\lfloor x\right\rfloor\pars{x^{2} + 1}
\,\right\vert_{\ 0}^{\ 2}}^{\ds{=\ 10}}\ -\
\int_{0}^{2}\lfloor x\rfloor\pars{2x}\dd x
\\[3mm] = &\
10\ -\ \underbrace{2\int_{0}^{1}\lfloor x\rfloor\,\dd x}_{\ds{=\ 0}}\ -\ \underbrace{2\int_{1}^{2}\lfloor x\rfloor x\,\dd x}_{\ds{=\ 3}} =
\bbox[10px,#ffa,border:1px groove navy]{7} \\ &
\end{align}
A: AS a hint:an example
$$\int_{2}^{7} t^2 \, d\lfloor t \rfloor = 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 135$$
For the former, $t↦⌊t⌋$ increases only at its jumps on [$2,7]$, which are precisely $t=3,4,5,6,7$. Since the jump sizes are identically 1,
