Area Under the Curve
Area between curve and axes
Grade 12
Question:
<p>The area between the curve \(y = 2x^4 - x^2\), the <em>x</em>-axis and the ordinates of the two minima of the curve is \(p/q\), then</p>
<p>(a) \(p + q = 127\)</p>
<p>(b) \(p - q = 113\)</p>
<p>(c) \(p < 10\)</p>
<p>(d) \(q > 10\)</p>
Step-by-Step Solution
Key Concept: First find the minima of y = 2x⁴ - x² by setting dy/dx = 0, then integrate |y| between these x-values. The curve dips below the x-axis between its minima, requiring careful handling of negative regions.
<p><strong>Step 1:</strong> Find the minima of y = 2x⁴ - x²</p><p>dy/dx = 8x³ - 2x = 2x(4x² - 1) = 0</p><p>Critical points: x = 0, x = ±1/2</p><p><strong>Step 2:</strong> Determine which are minima using d²y/dx² = 24x² - 2</p><p>At x = ±1/2: d²y/dx² = 24(1/4) - 2 = 4 > 0 (minima)</p><p>At x = 0: d²y/dx² = -2 < 0 (maximum)</p><p>The two minima occur at x = -1/2 and x = 1/2</p><p><strong>Step 3:</strong> Evaluate y at the minima</p><p>y(±1/2) = 2(1/16) - 1/4 = 1/8 - 2/8 = -1/8</p><p><strong>Step 4:</strong> Set up the area integral from x = -1/2 to x = 1/2</p><p>Since the curve is below the x-axis in this region:</p><p>A = ∫₍₋₁/₂₎^(1/2) |2x⁴ - x²| dx = ∫₍₋₁/₂₎^(1/2) (x² - 2x⁴) dx</p><p><strong>Step 5:</strong> Evaluate using symmetry (integrand is even)</p><p>A = 2∫₀^(1/2) (x² - 2x⁴) dx = 2[x³/3 - 2x⁵/5]₀^(1/2)</p><p>= 2[(1/24) - 2(1/160)] = 2[(1/24) - (1/80)]</p><p>= 2[(10 - 3)/240] = 2(7/240) = 7/120</p><p>∴ p/q = 7/120, so p = 7, q = 120</p>
Correct Answer: A,B,D