The given integral is:

\[
\int_1^\infty x e^{-x} \, dx
\]

To solve this, we can use **integration by parts**.

Let:
- \( u = x \), so \( du = dx \)
- \( dv = e^{-x} dx \), so \( v = -e^{-x} \)

Now applying the integration by parts formula:

\[
\int u \, dv = uv - \int v \, du
\]

Substitute the values:

\[
\int_1^\infty x e^{-x} \, dx = \left[ -x e^{-x} \right]_1^\infty + \int_1^\infty e^{-x} \, dx
\]

### Step 1: Evaluate \( \left[ -x e^{-x} \right]_1^\infty \)
At \( x = \infty \):
\[
-x e^{-x} \to 0 \quad (\text{because } e^{-x} \to 0 \text{ faster than } x \to \infty)
\]

At \( x = 1 \):
\[
-x e^{-x} = -e^{-1}
\]

Thus:
\[
\left[ -x e^{-x} \right]_1^\infty = 0 - (-e^{-1}) = e^{-1}
\]

### Step 2: Solve the remaining integral \( \int_1^\infty e^{-x} \, dx \)
\[
\int_1^\infty e^{-x} \, dx = \left[ -e^{-x} \right]_1^\infty
\]

At \( x = \infty \):
\[
-e^{-x} = 0
\]

At \( x = 1 \):
\[
-e^{-x} = -e^{-1}
\]

Thus:
\[
\int_1^\infty e^{-x} \, dx = 0 - (-e^{-1}) = e^{-1}
\]

### Final Answer:
Now add the two results together:
\[
e^{-1} + e^{-1} = 2e^{-1}
\]

Thus, the value of the integral is:

\[
\int_1^\infty x e^{-x} \, dx = \frac{2}{e}
\]

---

Would you like further clarification on any of the steps?

Here are some related questions:
1. What are other examples of integrals solved by integration by parts?
2. How do we handle improper integrals like this one in general?
3. Can we apply different techniques to evaluate integrals involving \(e^{-x}\)?
4. What happens if we change the limits of integration?
5. How does the convergence of \(e^{-x}\) affect integrals on an infinite interval?

**Tip**: Always check the behavior of terms at infinity when evaluating improper integrals.

Learn how to solve the improper integral ∫∞ 1 x e^(-x) dx using integration by parts. This step-by-step solution covers key calculus concepts like integration by parts and the behavior of exponential functions at infinity. Suitable for undergraduate calculus students.

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