Recently, I asked this question: Integration of f(x) where f(x) is x in binary, used as a decimal. Please see this question for more details regarding the exact operations.
Following the answer to my question, I began to think about how to think about this intuitively. Using @MatthewLeingang's answer, we see a Riemann sum method of finding this area. Rather than use the complicated math he used, I looked at it this way.
If I have $2$ rectangles, then my "depth" is $log_2(2) = 1$. Therefore, I will go out until the first place after the decimal point. My summation will be $0.5(0.0 + 0.1)$. With depth $2$, I have $0.25(0.0 + 0.01 + 0.1 + 0.11)$.
Continuing this on, we see that we practically just exhaust all the possible combinations given a depth. Now, when we add up the inside of those parentheses, we notice that the value at each place value is the same. For depth $2$, it is $0.25(0.22)$, and for depth $4$, it is $0.0625(0.0001 + ... + 0.1111) = 0.0625(0.8888)$, etc.
Now, notice that all of these can be simplified down: $0.0625(0.8888) = 0.5(0.1111) \approx 0.25(0.22) = 0.5(0.11)$. From here, it's rather easy to notice that the answer is probably just $0.055555555...$ or $\frac{1}{18}$.
Next, looking at base $3$, I'll skip ahead a little. Basically where I got to is that each place value, once simplified, has the digit $3$. Looking at our expansion of combinations that we did in the previous base, we see that each digit can be either $0, 1$, or $2$, each equally. So, our equation is $\frac{1}{3}(0.333333333...)$.
Expanding this to base $n$, our value of the integral ends up being $\frac{1}{n}(0.\underline{\frac{n(n-1)}{2}}$ $\underline{\frac{n(n-1)}{2}}$ $\underline{\frac{n(n-1)}{2}})...$ which simplifies to $0.\underline{\frac{(n-1)}{2}}$ $\underline{\frac{(n-1)}{2}}$ $\underline{\frac{(n-1)}{2}}...$. Note that here the terms carry. This definition obviously works for base $2$, but trying it out for base $10$, we get $0.\underline{\frac{9}{2}}$ $\underline{\frac{9}{2}}$ $\underline{\frac{9}{2}}...$, or $0.\underline{4.5}$ $\underline{4.5}$ $\underline{4.5}...$. This simplifies to $0.\underline{4}$ $\underline{9}$ $\underline{9}$ $\underline{9}$ $\underline{9}...$ or basically $0.5$ which, is the $\int_0^1f(x)dx$ where $f(x) = x_{10}$ or $x$ base $10$, which is just $x$.
Something that I believe is true here is that the area for a given base $n$ can be found as the arithmetic mean of the area for two other bases $n+a$ and $n-a$. For example, if we take $f(x)$ using base $6$, we can take the average of the base $2$ and base $10$, which we already know: $\frac{1}{18} + \frac{1}{2} = \frac{5}{9}$, which, I'm pretty sure, is the area under base $6$.
Sorry for the long backstory, but I did have a couple of questions to throw out there.
1) Is the integral even valid for bases higher than 10? @MarkS. on the previous question posted a picture that showed the basics of the graph. Using the formula that I mentioned before would give us values for these integrals, but I'm not sure how/if they could be represented.
2) Are there any other interesting patterns to be noted with this idea? Any applications elsewhere?
Looking back, this is a really long question, so thanks in advance!
EDIT: Turns out that for a given base $n$, $A = \frac{n-1}{18}$.
In the same spirit as the calculation for binary, for base $3$ the digits are $0,1,2$, so the average digit is $1$. Then in the $n$th place past the decimal you get a contribution of $\frac 1{10^n}$ and the integral is $\sum_{n=1}^\infty \frac 1{10^n}=\frac 19$. For any base $b$ below $10$, the average digit is $\frac {b-1}2$ and the sum will be $\frac {b-1}{18}$. This works whether you keep $b$ in the place it belongs or carry the leading digits of $b$ (when the digit is greater than $10$) to previous columns in the expansion.