Define $f(x)$ when $x$ $\in [0,1]$ as $x_2$ ($x$ base 2) considered as a decimal value.
Therefore, $f(0.75) = 0.11$, and $f(0.25) = 0.01$.
Compute $\int_0^1f(x)dx$.
To do this, I figured that the answer might be $0$ because integrals aren't defined by the value at a point, and the graph shows its discontinuity. I also tried using a program, and I got $0$. Is this the correct answer? Is there a better way of doing it?
Here's why zero can't be right. You can see $f(x) \geq 0.1$ on $[0.5,1.0)$.
So $\int_0^1 f(x)\,dx \geq (0.1)(0.5) = 0.05$.
Also, $f(x) \geq 0.01$ on $[0.25,0.5)$, $f(x) \geq 0.1$ on $[0.5,0.75)$, and $f(x) \geq 0.11$ on $[0.75,1)$. So $$ \int_0^1 f(x)\,dx \geq (0.01)(0.25) + (0.1)(0.25) + (0.11)(0.25) = (0.22)(0.25) = 0.055 $$ Keep going. Then try to generalize.
Let $\chi_n(x)$ be the $n$th digit of $x$ in its binary expansion. Notice that $$ \int_0^1 \chi_n(x)\,dx = \frac{1}{2} $$ That is, half of the numbers in $[0,1]$ have this digit zero, and half have it one. Also, $$ f(x) = \sum_{n=1}^\infty \frac{\chi_n(x)}{10^n} $$ Integrating term-by-term, we have $$ \int_0^1 f(x)\,dx = \sum_{n=1}^\infty \frac{1/2}{10^n} = \frac{1/20}{1-1/10} = \frac{1}{18} = 0.05555\dotsc $$ You will need to make sure that integrating the series term-by-term is justified. That should be OK because $0 \leq \chi_n(x) \leq 1$. By the Weierstrass $M$-test, the series converges uniformly.