Definite Integration
Properties and evaluation of definite integrals
Grade 12
Question:
<p>If \(f\) and \(g\) are two functions such that \(2f(1) = g(2) = 4\) and \(2f(9) = g(10) = 20\) and \(\int_0^2 (x^2 g(f(x^3+1))f'(x^3+1) - 3x^2)\,dx = 0\), then find the value of \(\int_4^{20} g^{-1}(x)\,dx\).</p>
Step-by-Step Solution
Key Concept: Use the given integral condition to establish a relationship between f and g, then apply integration by parts on the inverse function using the boundary values provided by the constraint equation.
<p><strong>Step 1: Analyze the given integral condition</strong></p><p>∫₀² (x²g(f(x³+1))f'(x³+1) - 3x²)dx = 0</p><p>This implies: ∫₀² x²g(f(x³+1))f'(x³+1)dx = ∫₀² 3x²dx = [x³]₀² = 8</p><p><strong>Step 2: Determine the relationship between f and g</strong></p><p>Substitute u = x³+1, so du = 3x²dx. When x=0, u=1; when x=2, u=9:</p><p>∫₁⁹ g(f(u))f'(u)·(du/3) = 8</p><p>∫₁⁹ g(f(u))f'(u)du = 24</p><p>By substitution v = f(u): dv = f'(u)du, and using f(1)=2, f(9)=10 (from given conditions):</p><p>∫₂¹⁰ g(v)dv = 24</p><p><strong>Step 3: Use integration by parts for inverse function</strong></p><p>For ∫₄²⁰ g⁻¹(x)dx, use: ∫g⁻¹(x)dx = xg⁻¹(x) - ∫g(t)dt where t = g⁻¹(x)</p><p>At x=4: g⁻¹(4) = 2 (since g(2)=4)</p><p>At x=20: g⁻¹(20) = 10 (since g(10)=20)</p><p>∫₄²⁰ g⁻¹(x)dx = [xg⁻¹(x)]₄²⁰ - ∫₂¹⁰ g(t)dt</p><p>= (20·10 - 4·2) - 24</p><p>= (200 - 8) - 24</p><p>= 192 - 24</p><p>∴ Answer: <strong>168</strong></p>
Correct Answer: 168