3points | 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.)

The improper integral converges and .

Step-by-step explanation:

First, I assume that the integral in question is

.

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

converges, by definition, if the following limit exist:

.

In this particular case we need to study the limit

.

In order to complete this task we calculate the integral . Then,

.

Substituting the above expression into the limit we have

because

.

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !