Area Under the Curve
Area bounded by parabola
Grade 12
Question:
<p>Let \(S(\alpha) = \{(x, y) : y^2 \leq x,\ 0 \leq x \leq \alpha\}\) and \(A(\alpha)\) is area of the region \(S(\alpha)\). If for a \(\lambda\), \(0 < \lambda < 4\), \(A(\lambda) : A(4) = 2 : 5\), then \(\lambda\) equals:</p>
<p>\(2\left(\dfrac{4}{25}\right)^{1/3}\)</p>
<p>\(2\left(\dfrac{2}{5}\right)^{1/3}\)</p>
<p>\(4\left(\dfrac{2}{5}\right)^{1/3}\)</p>
<p>\(4\left(\dfrac{4}{25}\right)^{1/3}\)</p>
Step-by-Step Solution
Key Concept: The region S(α) is bounded by the parabola y² = x, so its area is A(α) = 2∫₀^α √x dx = (4/3)α^(3/2). Use the ratio condition A(λ)/A(4) = 2/5 to set up an equation in terms of λ^(3/2).
<p><strong>Step 1:</strong> Identify the region S(α). The parabola y² = x opens rightward. For fixed x ∈ [0, α], y ranges from -√x to +√x. This is symmetric about the x-axis.</p><p><strong>Step 2:</strong> Calculate A(α). By symmetry: A(α) = 2∫₀^α √x dx = 2 · [⅔x^(3/2)]₀^α = (4/3)α^(3/2)</p><p><strong>Step 3:</strong> Find A(4). A(4) = (4/3)·4^(3/2) = (4/3)·8 = 32/3</p><p><strong>Step 4:</strong> Use the ratio condition. A(λ)/A(4) = 2/5 gives: [(4/3)λ^(3/2)] / [32/3] = 2/5</p><p><strong>Step 5:</strong> Simplify. (4λ^(3/2))/32 = 2/5 → λ^(3/2)/8 = 2/5 → λ^(3/2) = 16/5</p><p><strong>Step 6:</strong> Solve for λ. λ = (16/5)^(2/3) = [2⁴/5]^(2/3) = 2^(8/3)/5^(2/3) = (2^8)^(1/3)/(5^2)^(1/3) = ∛(256/25)</p><p>Alternatively: λ^(3/2) = 16/5 → λ² = (16/5)^(4/3) or compute directly: λ = (16/5)^(2/3) ≈ 2.297</p><p>If answer choices are provided, ∛(256/25) or simplified form matches option D.</p>
Correct Answer: D