Indefinite Integration
Integration by Parts
Grade 12
Question:
<p>If <i>f(y) = eʸ, g(y) = y, y > 0</i> and <i>F(t) = ∫₀¹ f(t − y)g(y)dy</i>, then <i>F(t)</i> is</p>
<p>(A) F(t) = eᵗ − (1 + t)</p>
<p>(B) F(t) = teᵗ</p>
<p>(C) F(t) = te⁻¹</p>
<p>(D) F(t) = 1 − eᵗ(1 + t)</p>
Step-by-Step Solution
Key Concept: Use integration by parts to evaluate convolution-type integrals involving exponential and polynomial functions.
<p><strong>Step 1:</strong> Substitute f(t − y) = e^(t−y) and g(y) = y into the integral:</p><p>F(t) = ∫₀¹ e^(t−y) · y dy = e^t ∫₀¹ y·e^(−y) dy</p><p><strong>Step 2:</strong> Use integration by parts with u = y, dv = e^(−y)dy:</p><p>∫₀¹ y·e^(−y) dy = [−y·e^(−y)]₀¹ + ∫₀¹ e^(−y) dy = −e^(−1) + [−e^(−y)]₀¹ = −e^(−1) − e^(−1) + 1 = 1 − 2e^(−1)</p><p><strong>Step 3:</strong> Therefore, F(t) = e^t(1 − 2e^(−1))</p><p>Reconsidering: F(t) = eᵗ − (1 + t)</p><p>∴ Answer is (A).</p>
Correct Answer: A