Question

Evaluate the value of the following definite integral using standard calculus integrations: \( \int_{0}^{1} x e^x \, dx \)

• A. \( e \)
• B. \( 1 \)
• C. \( e - 1 \)
• D. \( 0 \)

Hint:

Be careful when evaluating lower limits at zero, especially with exponential functions. While algebraic variables like \( x \) drop to 0, exponential terms evaluate to 1 (\( e^0 = 1 \)), which often changes the final result.

Answer 1

The Correct Option is B

Concept: When an integrand consists of a product of two distinct functions, apply the Integration by Parts rule: \[ \int u \, dv = uv - \int v \, du \] Choose the parts systematically using the ILATE priority rule (Inverses, Logarithms, Algebraics, Trigonometrics, Exponentials).

Step 1: Assign integration variables following the ILATE rule.
For the integrand \( x e^x \), choose the algebraic term as \( u \) and the exponential term as \( dv \):

• Let \( u = x \quad \Rightarrow \quad du = dx \)
• Let \( dv = e^x \, dx \quad \Rightarrow \quad v = e^x \)

Step 2: Apply the integration by parts formula.
Substitute these terms into the integration by parts formula: \[ \int x e^x \, dx = x e^x - \int e^x \, dx \] Evaluate the remaining integral: \[ \int x e^x \, dx = x e^x - e^x = e^x(x - 1) \]

Step 3: Evaluate the definite integral boundaries.
Apply the limits from \( 0 \) to \( 1 \) across the integrated expression: \[ \int_{0}^{1} x e^x \, dx = \left[ e^x(x - 1) \right]_{0}^{1} \] Evaluate the upper limit (\( x = 1 \)) and subtract the lower limit (\( x = 0 \)): \[ = \left( e^1(1 - 1) \right) - \left( e^0(0 - 1) \right) \] \[ = 0 - (1 \cdot (-1)) = 0 - (-1) = 1 \]