Is $\int_0^2 f(x) dx$ defined for $f(x)=x,x \ne 1$? Let $f(x)=x,x \ne 1$.

Is $$\int_0^2 f(x) dx$$ defined?

I'm currently a high school student, and we learn that the integral is the area under the graph. But in calculus textbooks and websites, I see some stuff about "continuity" and things being "integrable". So I was wondering if this simple case is considered "integrable" (because $\lim_{x\to1} f(x)$ is defined) or not (because it has a value "missing" altogether, rather than being discontinuous but still integrable like a step function).

Thanks!
Technically speaking, no, it isn't, since the function needs to be defined on the entire interval $[0,2]$ for the integral over $[0,2]$ to be well-defined. That said, you can define $f(1)$ however you want so that you have a function defined on $[0,2]$; if you do this, the integral is defined and its value is $2$.
A single bad point in the interval that is integrated has zero Lesbegue measure, so it has zero impact on the integral.   However, study of Lesbegue theory of measure and integration is normally done in a graduate level class on analysis of functions of real variables and you may not have been there yet.

If you limit yourself to the Riemann method of integration, you cannot define the upper and lower sums for that bad point, so the integral cannot be defined.