calculus - Why does the antiderivative of $frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$)

Why does $\log{x}$ represent the area below the graph of $f(x)=\frac{1}{x}$ from $1$ to $x$? What's so special about $1$ in this case?

Of course I understand that $\log{1}=0$ and I also understand that you cannot start at $x=0$ because $f(0)$ is not defined.

Still I can't get my head around of why it has to be $1$.

Also, this further implies that part of the antiderivative of $f(x)=\frac{1}{x}$ has to be negative (as part of the function of $\log{x}$ is negative). But why is this necessary?

The background of this question is that my calculus-book (Calculus, a complete course) starts with noticing that $f(x)=\frac{1}{x}$ is not an antiderivative of a polynomial function and then attempts to define a antiderivative which ends up being $\log{x}$. It does this all before even addressing the fundamental theorem of calculus or techniques of integration. They then simply define $log(1)$ to be $0$ without even knowing what that function really is yet. So I am stuck with kind of a circular reasoning where log(1)=0 because we defined it that way, but I don't understand why we define it that way.

When calculating areas under graphs by taking the limit of a sum (instead of by integration), you would start at $x=0$ right?

So in short: Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$). Why do we define it that way?

I am looking for some kind of deeper understanding; something still didn't "click". So some real background on what's going on here would be much appreciated.

Thanks!

Answer

What is "the" anti-derivative of $x\mapsto x+3$? Well, there isn't really a "the" here. For any constant $C$, the map $x\mapsto\frac12x^2+3x+C$ ia an anti-derivative. It looks like $x\mapsto \frac12x^2+3x$, obtained by picking $C=0$ plays a special role among the anti-derivatives, but it doesn't. Somebody else might rightfully describe the family of anti-derivatives of $x\mapsto x+3$ as $x\mapsto \frac12(x+3)^2+C$ (and in fact this can happen in a systematic way by using the substitution rule when integrating), again with $C$ an arbitrary constant, but unrelated to the $C$ used above. Of course, if we set $C=0$ in this description, we arrive at $x\mapsto (x+3)^2$, not the same function as $x\mapsto \frac12x^2+3x$. So is one of them "better" than the other? This may be the case depending on context (e.g., the second form tells us immediately where the apex of the parabola is), but there is no general notion of better.

So what is an anti-derivative of $f\colon(0,\infty)\to \Bbb R$, $x\mapsto \frac1x$? It is any function $F\colon(0,\infty)\to \Bbb R$ with $F'(x)=f(x)$ for all $x>0$. If we do not have a suitable function with this property already in our utility belt, we might simply define a specific one this way and give it a name. To get rid of the $C$, we rely on the definite integral, i.e., we fix some $a$ and define $F(x)=\int_a^xf(t)\,\mathrm dt$. This merely shifts the burden: What $a$ do we pick?
$a=0$ would be nice, but $0$ is not in the domain of $f$; and not even $\lim_{a\to0}\int_a^xf(t)\,\mathrm dt$ exists. We might even try $a=+\infty$, but alas, $\lim_{a\to+\infty }\int_a^xf(t)\,\mathrm dt$ doesn't exist either (but for other $f$ this might still be a viable choice; for example, in physics, this is the default way to pick the integration constant when defining potentials). So we better pick $a$ somewhere between $0$ and $\infty$. But as $f(x)>0$ for all $x$, this automatically implies $F(x)a$, i.e., we cannot achieve that $F$ is non-negative (or non-positive) throughout.

But does that mean $a=1$ is the best choice? Maybe $a=\pi$ woul dmake things particularly elegant?
Well, as above, this may depend on the context, but there are some overwhelming contexts:
By substituting $x\leftarrow \frac1cx$ (with $c>0$), we see that
$$F(b)=\int_{a}^bf(x)\,\mathrm dx=\frac1c\int_{ac}^{bc}f({\tfrac xc})\,\mathrm dx=\frac1c\int_{ac}^{bc}\frac1c{x}\,\mathrm dx=\int_{ac}^{bc}\frac1{x}\,\mathrm dx=F(bc)-F(ac)$$
or: $F(bc)=F(b)+F(ac)$. This formula becomes most beautiful (namely: symmetric) if we agree on $a=1$: $F(bc)=F(b)+F(c)$.
This is one compelling reason to define our "go-to" anti-derivative (which we decide to call $\ln$) as $\ln x:=\int_1^x\frac 1t\,\mathrm dt$.