Question

If

\[ \int x e^{-x} \, dx = M e^{-x} + C, \quad \text{where } C \text{ is an arbitrary constant, then } M \text{ is equal to:} \]

• A. \( -(1 + x) \)
• B. \( 1 + x \)
• C. \( -2x \)
• D. \( x^2 \)
• E. 2

Hint:

Use integration by parts for integrals involving a product of functions. Choose \( u \) as the polynomial term and \( dv \) as the exponential term to simplify the integration.

Answer 1

The Correct Option is A

We are given the integral: \[ \int x e^{-x} \, dx = M e^{-x} + C \] We are tasked with finding the value of \( M \).
Step 1: Use integration by parts to solve the integral. Recall the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \] Let \( u = x \) and \( dv = e^{-x} \, dx \). Then, \( du = dx \) and \( v = -e^{-x} \). 
Step 2: Apply the integration by parts formula:
\[ \int x e^{-x} \, dx = -x e^{-x} - \int -e^{-x} \, dx \] Simplify: \[ \int x e^{-x} \, dx = -x e^{-x} + \int e^{-x} \, dx \] The integral of \( e^{-x} \) is \( -e^{-x} \), so: \[ \int x e^{-x} \, dx = -x e^{-x} - e^{-x} + C \] Step 3: Factor out \( e^{-x} \):
\[ \int x e^{-x} \, dx = -(x + 1) e^{-x} + C \] Thus, comparing with the given equation \( \int x e^{-x} \, dx = M e^{-x} + C \), we see that:
\[ M = -(x + 1) \] 
Thus, the correct answer is option (A).