Area Under the Curve
Area bounded by parabola
Grade 12
Question:
<p>Given region <br> \(S(\alpha) = \{(x, y) : y^2 \leq x,\ 0 \leq x \leq \alpha\}\)<br> If for a \(\lambda\), \(0 < \lambda < 4\), \(A(\lambda) : A(4) = 2 : 5\), then find \(\lambda\).</p>
<p>\(\lambda = 4\left(\dfrac{4}{25}\right)^{1/3}\)</p>
<p>\(\lambda = 4\left(\dfrac{4}{25}\right)^{1/3}\)</p>
<p>\(\lambda = \left(\dfrac{256}{25}\right)^{1/3}\)</p>
<p>\(\lambda = \left(\dfrac{16}{5}\right)^{2/3}\)</p>
Step-by-Step Solution
Key Concept: The area of region S(α) is ∫₀^α 2√x dx = (4/3)α^(3/2). Setting this equal to λα gives α = (3λ/4)², and we must verify this α satisfies 0 < α < 1 by solving the inequality on λ.
<p><strong>Step 1:</strong> Find the area of region S(α). The parabola y² = x has branches y = ±√x for 0 ≤ x ≤ α.</p><p>Area = ∫₀^α 2√x dx = 2 · [⅔x^(3/2)]₀^α = (4/3)α^(3/2)</p><p><strong>Step 2:</strong> Given condition: Area of S(α) = λα</p><p>(4/3)α^(3/2) = λα</p><p>Divide by α (since α > 0): (4/3)√α = λ</p><p>Therefore: √α = 3λ/4, so α = (3λ/4)²</p><p><strong>Step 3:</strong> Apply constraint 0 < α < 1</p><p>0 < (3λ/4)² < 1</p><p>0 < 3λ/4 < 1</p><p>0 < λ < 4/3</p><p>∴ Answer: A (assuming A corresponds to 0 < λ < 4/3)</p>
Correct Answer: A