Definite Integration
Definite integral using functional equations
Grade 12
Question:
<p>As we have \(f'(x) = f(x)\), \(f(0) = 1\) and \(f(x) + g(x) = x^2\). Then \(\displaystyle\int_0^1 f(x)\cdot g(x)\,dx\) equals:</p>
<p>\(e - \dfrac{1}{2}e^2 - \dfrac{3}{2}\)</p>
<p>\(e + \dfrac{1}{2}e^2 + \dfrac{3}{2}\)</p>
<p>\(e - \dfrac{1}{2}e^2 + \dfrac{3}{2}\)</p>
<p>\(e + \dfrac{1}{2}e^2 - \dfrac{3}{2}\)</p>
Step-by-Step Solution
Key Concept: Since f'(x) = f(x) with f(0) = 1, we have f(x) = eˣ. Use the constraint f(x) + g(x) = x² to find g(x), then compute the definite integral using integration by parts or direct substitution.
<p><strong>Step 1:</strong> Solve f'(x) = f(x) with f(0) = 1.</p><p>This is a standard separable ODE: df/f = dx, giving ln|f| = x + C.</p><p>With f(0) = 1, we get C = 0, so <strong>f(x) = eˣ</strong></p><p><strong>Step 2:</strong> Find g(x) using f(x) + g(x) = x².</p><p>g(x) = x² - eˣ</p><p><strong>Step 3:</strong> Compute ∫₀¹ f(x)·g(x) dx = ∫₀¹ eˣ(x² - eˣ) dx</p><p>= ∫₀¹ eˣ·x² dx - ∫₀¹ e²ˣ dx</p><p><strong>Step 4:</strong> For ∫eˣ·x² dx, use integration by parts twice:</p><p>∫eˣ·x² dx = eˣ(x² - 2x + 2)</p><p>Evaluating from 0 to 1: [e(1 - 2 + 2)] - [1(0 - 0 + 2)] = e - 2</p><p><strong>Step 5:</strong> For ∫₀¹ e²ˣ dx = [e²ˣ/2]₀¹ = (e² - 1)/2</p><p><strong>Step 6:</strong> Combine results:</p><p>∫₀¹ f(x)g(x) dx = (e - 2) - (e² - 1)/2 = e - 2 - e²/2 + 1/2 = <strong>e - 3/2 - e²/2</strong></p><p>∴ Answer: <strong>e - (e² + 3)/2</strong> or equivalent form</p>
Correct Answer: A