Area Under the Curve
Area between curve and chord
Grade 12
Question:
<p>Using matrix multiplication, consider the quadratic equation \(4x^2 f(-1) + 4x f(1) + f(2) = 3x^2 + 3x\). It is given that this equation has three roots \(x = a, b, c\); hence it is an identity. Therefore \(f(-1) = \dfrac{3}{4}\), \(f(1) = \dfrac{3}{4}\) and \(f(2) = 0\), giving \(f(x) = \dfrac{4 - x^2}{4}\). Let point \(A\) be \((-2, 0)\) and \(B\) be \((2t, -t^2 + 1)\) and maximum value of \(f(x) = 1\) at \(x = 0\). Now, as \(AB\) subtends a right angle at the vertex \(V(0, 1)\), find the required area (in sq. units) given by \(A = \displaystyle\int_{-2}^{8} \left(\dfrac{4 - x^2}{4} + \dfrac{3x + 6}{2}\right) dx\).</p>
Step-by-Step Solution
Key Concept: The area between two curves is found by integrating the difference of the upper and lower functions. Here, the linear function (from line AB) forms the upper boundary while the parabola f(x) = (4-x²)/4 forms the lower boundary over [−2, 8].
<p><strong>Step 1:</strong> Identify the integrand structure. We have f(x) = (4−x²)/4 and the line contributes (3x+6)/2. The combined integrand is: (4−x²)/4 + (3x+6)/2 = (4−x²)/4 + (6x+12)/4 = (16+6x−x²)/4</p><p><strong>Step 2:</strong> Set up the definite integral: A = ∫₋₂⁸ [(16+6x−x²)/4] dx = (1/4)∫₋₂⁸ (16+6x−x²) dx</p><p><strong>Step 3:</strong> Find the antiderivative: (1/4)[16x + 3x² − x³/3]₋₂⁸</p><p><strong>Step 4:</strong> Evaluate at upper limit x=8: 16(8) + 3(64) − 512/3 = 128 + 192 − 170.667 = 149.333</p><p><strong>Step 5:</strong> Evaluate at lower limit x=−2: 16(−2) + 3(4) − (−8)/3 = −32 + 12 + 2.667 = −17.333</p><p><strong>Step 6:</strong> Compute the difference: (1/4)[149.333 − (−17.333)] = (1/4)(166.667) = 41.67</p><p>∴ Answer: <strong>41.67 sq. units</strong></p>
Correct Answer: 41.67