Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 12

Any integral with infinite bounds is an improper integral therefore this is an improper integral.
int_-infty^0 e^(3x) dx=
Substitute u=3x => du=3dx, u_l=3cdot(-infty)=-infty, u_u=3cdot0=0.
1/3int_-infty^0 e^udu=1/3 e^u|_-infty^0=1/3(e^0-lim_(u to -infty)e^u)=
1/3(1-0)=1/3
As we can see, the integral converges and its value is equal to 1/3.
The image below shows the graph of the function and area under it corresponding to the integral. We can see that as x goes to minus infinity the function converges to zero and it does so "very fast" (exponentially to be more specific). Therefore, it should be no surprise that the above integral is a convergent one.