Question

Ap Calculus B.C. This involves improper integrals. I need help so give me a detailed explanation with steps.

Answer 1

$\int ^{-1}_{-\infty}e^{5t}dt$

The first thing we are going to do is a substitution:

$\begin{gathered} u=5t \\ du=5dt\to dt=\frac{1}{5}du \end{gathered}$

We replace the substitution in the integral and solve the integral

$\begin{gathered} \int ^{-1}_{-\infty}e^u\cdot(\frac{1}{5}\cdot du) \\ \frac{1}{5}\int ^{-1}_{-\infty}e^u\cdot\mathrm{d}u \end{gathered}$

We know that the integral of an exponential is the same exponential, so:

$\frac{1}{5}\int ^{-1}_{\infty}e^u\cdot\mathrm{d}u=\frac{1}{5}e^u$

To evaluate the answer we must recover the substitution made from the beginning:

$\frac{1}{5}e^u=\frac{1}{5}e^{5t}$

Now, we evaluate in the interval from negative infinity to negative 1

$\begin{gathered} \int ^{-1}_{-\infty}e^{5t}dt=\frac{1}{5}e^{5t} \\ \int ^{-1}_{-\infty}e^{5t}dt=\frac{1}{5}e^{5(-1)}-\frac{1}{5}e^{5(-\infty)} \\ \int ^{-1}_{-\infty}e^{5t}dt=\frac{1}{5}e^{-5}-0 \end{gathered}$$\int ^{-1}_{-\infty}e^{5t}dt=\frac{1}{5}e^{-5}=0.00135$

This integral is convergent