Area Under the Curve
Area bounded by trigonometric curve
Grade 12
Question:
<p>Given \(g(x) = \cos x^2\) and \(f(x) = \sqrt{x}\), and the equation \(18x^2 - 9\pi x + \pi^2 = 0\) has roots \(\alpha = 6x - \pi\) and \(\beta = 3x + \pi\), the area (in sq. units) bounded by the curve \(y = (g \circ f)(x) = \cos x\) between \(x = \dfrac{\pi}{6}\) and \(x = \dfrac{\pi}{3}\) and \(y = 0\) is</p>
<p>\(\dfrac{1}{2}(\sqrt{3} - 1)\)</p>
<p>\(\dfrac{1}{2}(\sqrt{3} + 1)\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: The composition (g ∘ f)(x) = cos(√x)² = cos(x) simplifies the problem; recognize that definite integration of cos(x) directly gives the bounded area using the antiderivative sin(x).
<p><strong>Step 1:</strong> Identify the composition function.</p><p>Given g(x) = cos(x²) and f(x) = √x</p><p>(g ∘ f)(x) = g(f(x)) = cos((√x)²) = cos(x)</p><p><strong>Step 2:</strong> Set up the area integral.</p><p>Area bounded by y = cos(x), y = 0 between x = π/6 and x = π/3:</p><p>A = ∫[π/6 to π/3] cos(x) dx</p><p><strong>Step 3:</strong> Evaluate the definite integral.</p><p>A = [sin(x)][π/6 to π/3]</p><p>A = sin(π/3) − sin(π/6)</p><p>A = (√3/2) − (1/2)</p><p>A = (√3 − 1)/2 sq. units</p><p><strong>Note:</strong> The quadratic equation is extraneous information and should be ignored.</p><p>∴ Answer: A = <strong>(√3 − 1)/2</strong></p>
Correct Answer: A