For a real number \(x\) we denote by \([x]\) the largest integer less than or equal to \(x\).
Let \(n\) be a natural number. The integral \[\int_0^n [2^x]\, dx\] equals
\(\log_2 ((2^n-1)!)\);
\(n2^n-\log_2 ((2^n)!)\);
\(n2^n\);
\(\log_2 ((2^n)!)\),
where \(k! = 1\times 2 \times 3 \times \cdots \times k\) for a positive integer \(k\).
The function \([x]\) is called the floor function (for obvious reasons). It can also be written as \(\lfloor x \rfloor\).
You might wonder if there is a ceiling function, and indeed there is; \(\lceil x \rceil\) represents the smallest integer greater than or equal to \(x\).
The graph of \([2^x]\) looks roughly like this:
![Graph of [2 to the x], which has the shape of steps](/calculus-trig-log/r7106/images/graph.png)
We can see that, in the range \(0\le x\le n\), \([2^x]\) will take all integer values from \(1\) to \(2^n\). More precisely, we have \[[2^x]=k \quad \text{for} \quad \log_2 k \le x < \log_2 (k+1).\]
This means we can turn the integral into a sum of rectangles which equals \[(\log_2 2-\log_2 1)+2(\log_2 3-\log_2 2)+3(\log_2 4-\log_2 3)+\cdots+(2^n-1)(\log_2 2^n-\log_2(2^n-1)).\]
There is a lot of cancellation here, which leaves us with \[\begin{align*} -\log_2 1&{}-\log_2 2-\log_2 3-\cdots-\log_2(2^n-1)+(2^n-1)\log_2 2^n\\ &{}=2^n\log_2 2^n-(\log_2 1+\log_2 2+\log_2 3+\cdots+\log_2 2^n)\\ &{}=n2^n-\log_2((2^n)!). \end{align*}\]So the answer is (b).
An alternative way to think about the area is as horizontal strips instead of vertical strips.
The strip between height \(k-1\) and \(k\) has length \(n-\log_2 k\), and so this is its area, too.
Adding these up for \(k=1\) through to \(k=2^n\) gives the same answer.
The strip between height \(2^n-1\) and \(2^n\) has zero width and zero area.
