I got a problem this evening which I might well be able to solve, but am unable to confidently interpret the information it has given me. It goes as follows:
If $n\in\Bbb{N}$ and $[x]$ is the integer value of $x$, show that
$$\int_0^n[2^{x}]dx=n2^n-log_2((2^n)!)$$
I have emboldened the bit that I don't understand. Does it only want me to take integer values for $x$, i.e: what would the graph of $y=[2^x]$ look like?
On
$$\int_{0}^{n}\left \lfloor 2^x \right \rfloor = \sum_{k=1}^{2^n-1}{k\cdot( log_2(k+1)-log_2(k))}=\sum_{k=1}^{2^n-1}{k\cdot( \frac{\ln(k+1)}{\ln2}-\frac{\ln(k)}{\ln2})}=\sum_{k=1}^{2^n-1}{k\cdot( \frac{\ln(k+1)-\ln(k)}{\ln2})}=\frac{1}{\ln2}\cdot\sum_{k=1}^{2^n-1}{k\cdot(\ln(k+1)-\ln(k)})=\frac{1}{\ln2}\cdot\sum_{S=1}^{2^n-1}\sum_{k=1}^{S}{(-\ln(k)+\ln({k+1}))}=\frac{1}{\ln2}\sum_{S=1}^{2^n-1}{\ln(S+1)}=\sum_{i=1}^{2^n}{\log_{2}(i))}=\log_{2}(2^n!)$$
The notation means $[x]$ is the least integer not greater than $x$. Thus $[2.9] = 2$ and $[-2.9] = -3$ and $[3.0] = 3$.
This is a bit old fashioned; modern practice is to call this the "floor" function $\lfloor x \rfloor$.
So $[2^x]$ is $1$ until $x=1$, then $2$ until $x=\log_2 3$, then $3$ until $x=\log_2 4 = 2$ and so forth.
The integral becomes a discrete sum; don't forget that the domains of $x$ in which the integrand takes on each successive integer are not all the same length.