Area Under the Curve
Area bounded by curves and lines
Grade 12
Question:
<p>Let <em>g</em>(x) = cos x², <em>f</em>(x) = √x and α, β (α < β) be the roots of the quadratic equation 18x² − 9πx + π² = 0. Then, the area (in sq. units) bounded by the curve y = (g ∘ f)(x) and the lines x = α, x = β and y = 0 is</p>
<p>\(\dfrac{1}{2}(\sqrt{3}+1)\)</p>
<p>\(\dfrac{1}{2}(\sqrt{3}+\sqrt{2})\)</p>
<p>\(\dfrac{1}{2}(\sqrt{2}-1)\)</p>
<p>\(\dfrac{1}{2}(\sqrt{3}-1)\)</p>
Step-by-Step Solution
Key Concept: Recognize that (g ∘ f)(x) = cos(x) after simplification, then find roots of the quadratic to determine integration bounds, and integrate cos(x) over [α, β].
<p><strong>Step 1:</strong> Simplify the composite function.<br>g(x) = cos(x²), f(x) = √x<br>(g ∘ f)(x) = g(f(x)) = cos((√x)²) = cos(x)</p><p><strong>Step 2:</strong> Solve the quadratic 18x² − 9πx + π² = 0 for roots α and β.<br>Using the quadratic formula: x = [9π ± √(81π² − 72π²)]/(36) = [9π ± 3π]/36<br>α = (9π − 3π)/36 = π/6 and β = (9π + 3π)/36 = π/3</p><p><strong>Step 3:</strong> Set up the area integral.<br>Area = ∫[π/6 to π/3] cos(x) dx</p><p><strong>Step 4:</strong> Evaluate the integral.<br>∫cos(x) dx = sin(x) |[π/6 to π/3]<br>= sin(π/3) − sin(π/6)<br>= (√3/2) − (1/2)<br>= (√3 − 1)/2</p><p>∴ Answer: D</p>
Correct Answer: D