so there's this problem my calc textbook uses and the solution is slightly unclear to me. So it asks to evaluate the integral of e^(-x^2) from an interval from 0 to infinity. Here's how it goes about this
In the end, it only evaluates the integral from 1 to infinity. What is not clear to me is that since it has proven that it is convergent from 1 to infinity, it follows that it is also convergent from 0 to infinity (by comparison theorem). I figure that it ends up graphing both f(x), g(x) as a visual aid to see why this is so. But without the graphs shown, would the solution still be valid? This is how it defines the comparison theorem
Clearly, both functions have to have the same interval in order to say that the other is convergent without having to evaluate both of them.
It might prove useful to elaborate on the text's argument. Firstly,
$$\int_0^\infty e^{-x^2} dx = \int_0^1 e^{-x^2} dx + \int_1^\infty e^{-x^2} dx $$
The first is clearly what the text calls an "ordinary" definite integral, i.e. it has a finite value. How can we necessary conclude this? Well, it is not an improper integral of either "type:" those being the types that are over rays, infinite in length, and those that encounter bad behavior or other singularities. Since the interval over which we integrate is $[0,1]$, it has finite length and is not the first kind of improper integral. The graph of the integrand on this interval should make clear that the function is well-behaved (for example, it does not "blow up" to infinity as $1/x$ would as $x \to 0$).
More explicitly, the "well-behaved"-ness is that the function is bounded and continuous, and is integrated over an interval of finite length. These three conditions ensure that the integral converges to a finite value.
What we conclude thus is that $\int_0^1 e^{-x^2} dx$ is finite. What is its value? Who knows, but it's something finite, and in terms of convergence for now that's all we really care about. Let us call this finite value $k$ because we can.
What about the second integral? As it happens, since $e^x$ is a monotone increasing function, and $x \ge 1$ when $x \in [1,\infty)$ (the interval of integration),
$$x^2 \ge x \implies -x^2 \le -x \implies e^{-x^2} \le e^{-x}$$
By the comparison theorem, we can thus claim the following:
$$\int_1^\infty e^{-x^2} dx \le \int_1^\infty e^{-x} dx$$
The latter is a simple integral to evaluate, and evaluates to $1/e$ as described.
Through these facts, we conclude, since two finite values always sum to another finite one:
$$\begin{align} \int_0^\infty e^{-x^2} dx &= \int_0^1 e^{-x^2} dx + \int_1^\infty e^{-x^2} dx \\ &\le k + \int_1^\infty e^{-x} dx \\ &= k + \frac 1 e \\ &< \infty \end{align}$$
You are correct in that convergence on $[1,\infty)$ does not ensure convergence on $[0,\infty)$. A slightly contrived example of this would be
$$f(x) = \left\{\begin{matrix} e^{-x^2} & x \ge 1\\ \frac 1 x & x \le 1 \end{matrix}\right.$$
For this function, clearly we have convergence on $[1,\infty)$ as established, but to evaluate the integral on $[0,\infty)$ you will have to contend with $\int_0^1 \frac 1 x dx$ in the evaluation, which is obviously infinite.
This fact is why we split up the integral between $[0,1]$ and $[1,\infty)$. We verify the convergence on the first interval by one method, and convergence on the second interval by another, and thus their sum converges as well.