Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Answer 1

Answer:

$\int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}$  

Step-by-step explanation:

Assuming this integral:

$\int_{-\infty}^0 5 e^{60x} dx$

We can do this as the first step:

$5 \int_{-\infty}^0 e^{60x} dx$

Now we can solve the integral and we got:

$5 \frac{e^{60x}}{60} \Big|_{-\infty}^0$

$\int_{-\infty}^0 5 e^{60x} dx = \frac{e^{60x}}{12}\Big|_{-\infty}^0 = \frac{1}{12} [e^{60*0} -e^{-\infty}]$

$\int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}$  

So then we see that the integral on this case converges amd the values is 1/12 on this case.