Question

\[ \int_0^\infty x e^{-x} dx = \text{...............} \] (round off to the nearest integer)

Hint:

The integral \( \int_0^\infty x e^{-x} dx \) can be evaluated using integration by parts. The result is \( 1 \).

Answer 1

We need to evaluate the integral: \[ I = \int_0^\infty x e^{-x} dx \] Step 1: Use integration by parts.
Let \( u = x \) and \( dv = e^{-x} dx \). Then, \[ du = dx \quad \text{and} \quad v = -e^{-x} \] Now apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substitute: \[ I = \left[ -x e^{-x} \right]_0^\infty + \int_0^\infty e^{-x} dx \] Step 2: Evaluate the boundary terms.
At \( x = \infty \), \( x e^{-x} \to 0 \), and at \( x = 0 \), \( x e^{-x} = 0 \). So the boundary term evaluates to 0: \[ \left[ -x e^{-x} \right]_0^\infty = 0 \] Step 3: Evaluate the remaining integral.
The remaining integral is: \[ \int_0^\infty e^{-x} dx = \left[ -e^{-x} \right]_0^\infty = 1 \] Final Answer: \[ I = 1 \] Final Answer: \[ \boxed{1} \]