A strange-looking identity How do I prove for a continuous function $f$ that
$$
\int_1^x\lfloor{u}\rfloor\lfloor{u+1}\rfloor f(u)du=2\sum_{i=1}^{\lfloor{x}\rfloor}i\int_i^xf(u)du\ ?
$$
This is what I've come up with so far: 
$$
\int_1^x\lfloor{u}\rfloor\lfloor{u+1}\rfloor f(u)du=\sum_{i=1}^{\lfloor{x}\rfloor-1}i(i+1)\int_i^{i+1} f(u) du+\lfloor x\rfloor \lfloor x+1\rfloor \int_{\lfloor x\rfloor}^{x}f(u)du.
$$
It doesn't seem to be of any help though.
 A: It is easy. Start from the RHS to get
$$2\left(1\cdot\int_1^xf(u)du+2\cdot\int_2^xf(u)du+\cdots+\lfloor x\rfloor\int_{\lfloor x\rfloor}^x f(u)du\right)$$
$$\small =2\left(\int_1^2 f(u)du+(1+2)\int_2^3 f(u)du+(1+2+3)\int_3^4f(u)du+...+(1+2+...+\lfloor x\rfloor)\int_{\lfloor x\rfloor}^x f(u)du\right),$$
which equals the LHS, since $2(1+2+\cdots+\lfloor u\rfloor)=\lfloor u\rfloor \lfloor u+1\rfloor$.
A: Hint. For any $u\ge0$ and $a\ge u$,
$$1+2+\cdots+n=\frac{n(n+1)}{2} ~~\implies~~ \frac{\lfloor u\rfloor \big(\lfloor u\rfloor+1\big)}{2}=\sum_{0\le k\le u}k=\sum_{k=1}^a kH(u-k),$$
where (for our purposes) we define
$$H(x)=\begin{cases}1 & x\ge0 \\ 0 & x<0.\end{cases}$$
Sidequest: for arbitrary integrable $g$ and real $m\in[a,b]$, how can we simplify $\int_a^b g(x)H(x-m)dx$?
A: This may be seen as the result of a summation by parts, aka a discrete Abel transform.
To wit, start from the formula in your post and write every integral $\displaystyle\int_i^{i+1}f$ as $S_i-S_{i+1}$, where $S_i=\displaystyle\int_i^xf$. Thus, introducing $k=\lfloor x\rfloor$, you are interested in
$$
T_k=\sum_{i=1}^{k-1}i(i+1)(S_i-S_{i+1})+k(k+1)S_k.
$$
One sees that
$$
T_k=\sum_{i=1}^{k-1}i(i+1)S_i-\sum_{i=1}^{k-1}i(i+1)S_{i+1}+k(k+1)S_k=\sum_{i=1}^{k}i(i+1)S_i-\sum_{i=1}^{k-1}i(i+1)S_{i+1}.
$$
For every $2\leqslant i\leqslant k$, the coefficient of $S_i$ is $i(i+1)-(i-1)i=2i$, and the coefficient of $S_1$ is $1\cdot2=2$. QED.
