Review question

# Can we find this integral involving the floor function? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7106

## Suggestion

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

1. $\log_2 ((2^n-1)!)$;

2. $n2^n-\log_2 ((2^n)!)$;

3. $n2^n$;

4. $\log_2 ((2^n)!)$,

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?