Answer:
The improper integral converges and [tex]\int_0^{-\infty} e^{7x}dx = -\frac{1}{7}[/tex].
Step-by-step explanation:
First, I assume that the integral in question is
[tex]\int_0^{-\infty} e^{7x}dx[/tex].
Now, the integral is improper because, at least, one of the limits is [tex]\pm\infty[/tex]. We need to recall that an improper integral
[tex]\int_0^{-\infty} f(x)dx[/tex]
converges, by definition, if the following limit exist:
[tex]\lim_{A\rightarrow -\infty} \int_0^A f(x)dx = \int_0^{-\infty} f(x)dx[/tex].
In this particular case we need to study the limit
[tex]\lim_{A\rightarrow -\infty} \int_0^A e^{7x}dx[/tex].
In order to complete this task we calculate the integral [tex]\int_0^A e^{7x}dx[/tex]. Then,
[tex]\int_0^A e^{7x}dx = \frac{e^{7x}}{7}\Big|_0^A = \frac{e^{7A}}{7} - \frac{1}{7}[/tex].
Substituting the above expression into the limit we have
[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7} - \frac{1}{7} = - \frac{1}{7}[/tex]
because
[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7}=0[/tex].
