An improper integral of Type I is an integral whose limits of integration include $\infty$ or $-\infty$, or both. Remember that $\infty$ is a process (keep going and never stop), not a number!
With this in mind, we define

When an integral runs from $-\infty$ to $\infty$, we have to break the integral into two pieces:$$\int_{-\infty}^\infty f(x)\, dx = \int_{-\infty}^a f(x) \,dx + \int_a^\infty f(x) \,dx,$$where we can choose any number for the break point $a$. (Zero is often convenient.) To evaluate the limits as $t \to \infty$ or $t \to -\infty$, we might need to use L'Hospital's rule.

On Convergence

We'll say that $\displaystyle\int_a^\infty f(x)\,dx$
converges if $\displaystyle\lim_{t\to\infty}\int_a^t f(x)\,dx$
exists, and that the improper integral diverges if the limit doesn't exist.

We define convergence/diverge of $\displaystyle\int_{-\infty}^b f(x)\,dx$ similarly in terms of the existence of its corresponding limit.

For
$\displaystyle\int_{-\infty}^\infty f(x)\,dx$ to converge we need both
$\displaystyle\lim_{t\to\infty}\int_{a}^t f(x)\,dx$ and
$\displaystyle\lim_{t\to-\infty}\int_{t}^a f(x)\,dx$ to exist.

We'll be talking a lot more about convergence and divergence when we get to sequences and series.

The following video explains Type I improper integrals and works out a number of examples.