Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 5

The integral does not have infinite bounds and the function is well defined over the whole interval of integration so there is no need to use limits. Therefore, the integral is not improper.
int_0^2 e^-x dx=
Substitute u=-x => du=-dx, u_l=0, u_u=-2.
u_l and u_u denote lower and upper bound of integration.
-int_0^-2 e^u du=int_-2^0 e^u du=e^u|_-2^0=e^0-e^(-2)=1-e^-2
As we can see there was no need to use limits for calculating the integral.
The image below shows the graph of the function. We can see from the image that the function is not only defined but continuous over the whole interval of integration. In fact, domain of exponential function is set of all real numbers (can be extended to all complex numbers).