Definite Integration
Convolution integrals
Grade 12
Question:
<p>If \(f(y) = e^y\), \(g(y) = y;\, y > 0\) and \(F(t) = \int_0^t f(t-y)g(y)\, dy\), then</p>
<p>\(F(t) = 1 - e^{-t}(1+t)\)</p>
<p>\(F(t) = e^t - (1+t)\)</p>
<p>\(F(t) = te^t\)</p>
<p>\(F(t) = te^{-t}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a convolution integral. Substitute u = t - y to transform the integral into a standard form that can be evaluated using integration by parts or pattern recognition.
<p><strong>Step 1:</strong> Write the given integral: F(t) = ∫₀ᵗ e^(t-y) · y dy</p><p><strong>Step 2:</strong> Substitute u = t - y, so y = t - u and dy = -du. When y = 0, u = t; when y = t, u = 0.</p><p><strong>Step 3:</strong> F(t) = ∫ₜ⁰ e^u(t - u)(-du) = ∫₀ᵗ e^u(t - u) du</p><p><strong>Step 4:</strong> Expand: F(t) = ∫₀ᵗ te^u du - ∫₀ᵗ ue^u du</p><p><strong>Step 5:</strong> Evaluate first integral: ∫₀ᵗ te^u du = t(e^t - 1)</p><p><strong>Step 6:</strong> For second integral, use integration by parts: ∫ue^u du = ue^u - e^u. So ∫₀ᵗ ue^u du = (te^t - e^t) - (0 - 1) = te^t - e^t + 1</p><p><strong>Step 7:</strong> F(t) = t(e^t - 1) - (te^t - e^t + 1) = te^t - t - te^t + e^t - 1 = e^t - t - 1</p><p>∴ Answer: D (which should be F(t) = e^t - t - 1)</p>
Correct Answer: D