If ∫₀¹ eˣ (x − α) dx = 0, then α is:

If ∫₀¹ eˣ (x − α) dx = 0, then α is:

• A. e − 1
• B. (e − 2)/(e − 1)
• C. e/(e − 1)
• D. 1/(e − 1)

Answer: B) (e − 2)/(e − 1)

Explanation: ∫₀¹ eˣ (x − α) dx = ∫₀¹ x eˣ dx − α ∫₀¹ eˣ dx. ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = e − (e − 1) = 1. ∫₀¹ eˣ dx = e − 1. So 1 − α(e − 1) = 0 → α = 1/(e − 1)? But check integration by parts: u=x, dv=eˣ dx, du=dx, v=eˣ. [x eˣ]₀¹ = e − 0 = e. ∫₀¹ eˣ dx = e − 1. So ∫₀¹ x eˣ dx = e − (e − 1) = 1. Correct. Then 1 − α(e−1)=0 → α = 1/(e−1). That's option 3. But we verify: α = (e−2)/(e−1) would give 1 − (e−2) = 3−e, not zero. So α = 1/(e−1). Option index 3.