Question

Let $[x]$ denote the largest whole number that is less than or equal to $x$. For example, $[-\pi] = -4$. A function $f(x)$ is defined as follows; if $0 < x < 2$ then

$f(x) = (\frac{3}{4})^{[\log_2(x)]}$

and $f(x) = 0$ otherwise. Note that, for example, $f(\frac{1}{2}) = \frac{4}{3}$. The value of $\int_0^2 f(x) \, dx$ is

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (C)

3

Answer 2

The integral $\int_0^2 f(x) \, dx$ is evaluated by splitting the interval $(0, 2)$ into sub-intervals where $f(x)$ is constant. This is determined by the floor function $[\log_2(x)]$.

For $2^n \le x < 2^{n+1}$, $[\log_2(x)] = n$, and $f(x) = (3/4)^n$.

The integral over such an interval is $\int_{2^n}^{2^{n+1}} (3/4)^n \, dx = (3/4)^n (2^{n+1} - 2^n) = (3/4)^n (2^n) = (3/2)^n$.

The integral from $0$ to $2$ sums these contributions for $n = 0, -1, -2, \dots$

$\int_0^2 f(x) \, dx = \sum_{n=0}^{-\infty} (3/2)^n = \sum_{k=0}^{\infty} (3/2)^{-k} = \sum_{k=0}^{\infty} (2/3)^k$.

This is an infinite geometric series: $1 + 2/3 + (2/3)^2 + \dots$.

The sum is $\frac{1}{1 - 2/3} = \frac{1}{1/3} = 3$.