Why do integrals "start" at 0?

When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0).

This is not true. In certain situations it may be the case, but not generally. I think the reason this confusion arises is that a common problem given to calculus students is to find the antiderivative of a polynomial, e.g.,

$$\int x^3 +2x \, dx = \frac{1}{4}x^4 + x^2 + C$$

and in this case, if we set $C = 0$ we get

$$\frac{1}{4}x^4 + x^2$$

which is the same as $$\int_0^x u^3 +2u \, du = \frac{1}{4}u^4 + u^2 \big|_{u=0}^{u=x}.$$

This will work whenever the form that the antiderivative $F$ of $f$ you get takes satisfies $F(0) = 0$. But in general, setting $C = 0$ will not get you the integral $\int_0^x f(t) \, dt$. For example, if you take $f(x) = e^x$ then $$\int e^x \, dx = e^x + C$$ but setting $C = 0$ gives you $e^x$, which is not the same as $$\int_0^x e^t \, dt = e^x - 1.$$

Note that "setting "C = 0" in the expression for the antiderivative" is not actually a well-defined operation. Different methods of antidifferentiation can give you different expressions when you set $C = 0$. It is important to remember that there is no single antiderivative, and no canonical way of writing it. $\int 2x \, dx = x^2 + 3 + C$ is just as valid as $\int 2x \, dx = x^2 + C$.

Integrals don't always start at $0$. Let's start from definite integrals and move to the indefinite integrals you asked about. We know $\int_a^b f(x)dx $ gives the area under the curve between $a$ and $b$. If $f(x)$ has an antiderivative $F(x)$, the Fundamental Theorem of Calculus tells us that $\int_a^b f(x) = F(b) - F(a)$. Thus, if we want the area under the curve from $a$ to $b$, we compute $F(b) - F(a)$. If we want the area under the curve from $0$ to $b$, we compute $F(b) - F(0)$. $F(b) - F(0)$, as a function of $b$, gives us the area from $0$ to $b$.

Now there are a lot of "natural" functions where $F(0) = 0$ (e.g., functions like $x^2$ or $\sin x$), so to $F(b)$ gives the area from $0$ to $b$. But that won't be the case if $F(0) \neq 0$.

The above should make it clear that there's no reason $0$ is special -- if you want the area from $-1$ to $b$ as a function of $b$, just use $F(b) - F(-1)$.

Well, integrals (by which I think you mean anti derivatives) don't always start at $0$. Indeed, if a function $f=f(x)$ is continuous in some interval $I=[a,b]$, then it has an anti derivative given by $$\int_c^x{f(t)dt},$$ where $c\in I$. The $0$ is usually chosen for $c$ only as a matter of convenience. A well-known function defined by an antiderivative that "starts" from (note that the notion of starting from for antiderivatives should not be taken too literally since the function is defined even for $x<c\in I$) $1$, not $0$ as usual, is the logarithm function $$\log x=\int_1^x{\frac{1}{t}dt}$$ defined for all $x>0$.