The integral shown in the image is:

$\int_{-\infty}^{-1} e^x \, dx$

Step-by-step solution:

1. Integral of $e^x$: The indefinite integral of $e^x$ is $e^x$ itself.

   $\int e^x \, dx = e^x + C$

2. Evaluate the definite integral: Now, we apply the limits of integration from $-\infty$ to $-1$.

   $\int_{-\infty}^{-1} e^x \, dx = \left[ e^x \right]_{-\infty}^{-1}$

3. Substitute the upper limit $x = -1$:

   $e^{-1} = \frac{1}{e}$

4. Substitute the lower limit $x = -\infty$: As $x \to -\infty$, $e^x \to 0$.

   $e^{-\infty} = 0$

5. Final Calculation: Now, subtract the lower limit from the upper limit:

   $\int_{-\infty}^{-1} e^x \, dx = \left( \frac{1}{e} - 0 \right) = \frac{1}{e}$

Thus, the value of the integral is:

$\boxed{\frac{1}{e}}$

Let me know if you would like more details or explanations!

Here are five related questions you might explore:

1. What is the integral of $e^x$ from $0$ to $1$?
2. How do you solve $\int_{-\infty}^{\infty} e^{-x^2} \, dx$?
3. What happens to an integral when the limits involve infinity?
4. How can you evaluate integrals involving other exponential functions, like $e^{2x}$?
5. How would you approach the integral $\int_{-1}^{1} e^x \, dx$?

Tip: When evaluating improper integrals, always ensure that the function converges, especially when limits involve infinity.