Question

3 points | Previous Answers LarCalcET6 8.8.012. Ask Your Teacher My Notes Question Part Points Submissions Used Explain why the integral is improper. 0 e7x dx −[infinity] At least one of the limits of integration is not finite. The integrand is not continuous on (-[infinity], 0]. Determine whether it diverges or converges. converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)

Answer 1

Answer:

The improper integral converges and $\int_0^{-\infty} e^{7x}dx = -\frac{1}{7}$.

Step-by-step explanation:

First, I assume that the integral in question is

$\int_0^{-\infty} e^{7x}dx$.

Now, the integral is improper because, at least, one of the limits is $\pm\infty$. We need to recall that an improper integral

$\int_0^{-\infty} f(x)dx$

converges, by definition, if the following limit exist:

$\lim_{A\rightarrow -\infty} \int_0^A f(x)dx = \int_0^{-\infty} f(x)dx$.

In this particular case we need to study the limit

$\lim_{A\rightarrow -\infty} \int_0^A e^{7x}dx$.

In order to complete this task we calculate the integral $\int_0^A e^{7x}dx$. Then,

$\int_0^A e^{7x}dx = \frac{e^{7x}}{7}\Big|_0^A = \frac{e^{7A}}{7} - \frac{1}{7}$.

Substituting the above expression into the limit we have

$\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7} - \frac{1}{7} = - \frac{1}{7}$

because

$\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7}=0$.