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 \(k\).

Can we sketch the function? What does the integral we need to find look like now?