Definite Integration
Definite Integration involving Determinants
Grade 12
Question:
<p>Let \(f(x) = \begin{vmatrix} 2\cos^2 x & \sin 2x & -\sin x \\ \sin 2x & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{vmatrix}\). Then the value of \(\displaystyle\int_0^{\pi/2} [f(x) + f'(x)]\,dx\) is</p>
<p>(1) \(\pi\)</p>
<p>(2) \(\pi/2\)</p>
<p>(3) \(2\pi\)</p>
<p>(4) \(3\pi/2\)</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant f(x) simplifies to a manageable form, and use the property that ∫[f(x) + f'(x)]dx = [f(x)]₀^(π/2) + ∫f(x)dx by integration by parts insight, or compute f(x) explicitly first to find a pattern.
<p><strong>Step 1: Simplify the determinant f(x)</strong></p><p>Expand the determinant:</p><p>f(x) = 2cos²x(0 + cos²x) - sin2x(0 - sin x·cos x) - sin x(-sin 2x·cos x - 2sin²x·sin x)</p><p>= 2cos⁴x + sin2x·sin x·cos x + sin x(sin 2x·cos x + 2sin³x)</p><p>= 2cos⁴x + sin2x·sin x·cos x + sin x·cos x·sin 2x + 2sin⁴x</p><p>= 2cos⁴x + 2sin⁴x + 2sin 2x·sin x·cos x</p><p>= 2(cos⁴x + sin⁴x) + 2·2sin x cos x·sin x·cos x</p><p>= 2(cos⁴x + sin⁴x) + 4sin²x·cos²x</p><p>= 2[(cos²x + sin²x)² - 2sin²x·cos²x] + 4sin²x·cos²x</p><p>= 2[1 - 2sin²x·cos²x] + 4sin²x·cos²x</p><p>= <strong>2</strong></p><p><strong>Step 2: Integrate [f(x) + f'(x)]</strong></p><p>Since f(x) = 2 (constant), f'(x) = 0</p><p>∫₀^(π/2) [f(x) + f'(x)]dx = ∫₀^(π/2) [2 + 0]dx = 2x|₀^(π/2) = 2(π/2) - 0</p><p>∴ Answer: <strong>π</strong></p>
Correct Answer: B