$$\int_0^4 \lfloor x/2 \rfloor \ d(x-\lfloor x \rfloor)$$
I don't get how we convert the given differential element into normal dx differential element.
I plotted the graphs of $$\lfloor x/2 \rfloor$$ and $$x-\lfloor x \rfloor$$ and tried to integrate along the graph of differential element function according to it answer comes to 0.
But the answer comes to 2
Just help in conversion of differential element the rest I can do
$x-\lfloor x\rfloor$ is the fractional part function, $\{x\}$. For $x\in[m,m+1),\{x\}=x-m$, where $m\in\Bbb Z$.
The integrand is $0$ in $[0,2)$ and $1$ in $[2,4)$. Your integral is$$\int_2^3d(x-2)+\int^4_3d(x-3)=2$$