The given integral is:

\[
\int_0^\infty x^0 e^{-x} \, dx
\]

First, observe that \(x^0 = 1\) for all \(x\), so the integral simplifies to:

\[
\int_0^\infty e^{-x} \, dx
\]

This is a standard integral, which can be computed as follows:

\[
\int_0^\infty e^{-x} \, dx = \left[-e^{-x}\right]_0^\infty
\]

At \(x = \infty\), \(e^{-\infty} = 0\), and at \(x = 0\), \(e^0 = 1\), so:

\[
\left[-e^{-x}\right]_0^\infty = 0 - (-1) = 1
\]

Thus, the value of the integral is:

\[
\boxed{1}
\]

Would you like further details or explanations on any step?

Here are 5 related questions for further practice:
1. What is \(\int_0^\infty x^n e^{-x} \, dx\) for any positive integer \(n\)?
2. How does the Gamma function generalize this type of integral?
3. What is the value of \(\int_0^\infty e^{-ax} \, dx\) for any positive constant \(a\)?
4. How would you compute \(\int_0^1 e^{-x} \, dx\)?
5. What happens if the upper bound of the integral changes to a finite number instead of infinity?

**Tip:** For integrals involving \(e^{-x}\) and powers of \(x\), the Gamma function \(\Gamma(n)\) can be a helpful tool to generalize the solution.

This guide explains how to solve the integral ∫_0^∞ e^(-x) dx, including key calculus concepts like definite integrals and exponential decay functions. Perfect for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation