Is there a curve for which the area underneath between n and n+1 is exactly double the area underneath the curve between n+1 and n+2?

Question

For example: $$\int_x^{x+1} f(x)\ dx = \frac{1}{2^x}$$ So for x = 0, the integral would be from 0 to 1 and would return 1. For x = 1, the integral would be from 1 to 2 and would return 0.5 . And so on...

Answer 1

This could be a solution:

First off apply the fundamental theorem of calculus

$$ \frac{{\rm d}}{{\rm d}x}\int_{x}^{x+1}{\rm d}z~f(z) = f(x+1) - f(x) $$

therefore, the function $f$ should satisfy

$$ f(x+1) - f(x) = -\frac{\ln 2}{2^x} \tag{1} $$

A solution for this is

$$ f(x) = \ln 4 - 2^{1-x}(2^x - 1)\ln 2 \tag{2} $$

The term $\ln 4$ is actually arbitrary, I just selected that number to keep $f(x) > 0$