To solve the integral:

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

Step-by-Step Solution:

1. Identify the integral: The integral is of the form: $\int_{0}^{\infty} e^{-x} \, dx$

2. Antiderivative of $e^{-x}$: The indefinite integral of $e^{-x}$ is: $\int e^{-x} \, dx = -e^{-x} + C$

3. Apply limits: We now evaluate: $\int_{0}^{\infty} e^{-x} \, dx = \left[ -e^{-x} \right]_{0}^{\infty}$

   • At $x = \infty$: $-e^{-x} \to 0$, because $e^{-\infty} = 0$.
   • At $x = 0$: $-e^{-x} = -e^{0} = -1$.

4. Combine results: $\int_{0}^{\infty} e^{-x} \, dx = 0 - (-1) = 1$

Final Answer:

$\boxed{1}$

Would you like further clarification or a more detailed explanation?

Here are 5 related questions to deepen your understanding:

1. What is the integral of $e^{-kx}$ where $k > 0$ from 0 to $\infty$?
2. How does the exponential decay factor $e^{-x}$ affect the convergence of the integral?
3. What is the relationship between this integral and the Gamma function?
4. How does this integral relate to probability distributions (e.g., the exponential distribution)?
5. Can improper integrals like this always be evaluated analytically?

Tip: Improper integrals often converge when the integrand decreases rapidly to zero as $x \to \infty$.