Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Let <em>f(x)</em> and <em>g(x)</em> be continuous, positive functions such that \(f(-x) = g(x) - 1\), \(f(x) = \dfrac{g(x)}{g(-x)}\) and \(\displaystyle\int_{-20}^{20} f(x)\,dx = 2020\), then the value of \(\displaystyle\int_{-20}^{20} \dfrac{f(x)}{g(x)}\,dx\) is:</p>
<p>(a) 1010</p>
<p>(b) 1050</p>
<p>(c) 2020</p>
<p>(d) 2050</p>
Step-by-Step Solution
Key Concept: Use the two functional equations f(-x) = g(x) - 1 and f(x) = g(x)/g(-x) to derive that g(x)·g(-x) = g(x)/(g(x)-1), which leads to g(x) + g(-x) = 1. Then exploit symmetry properties of the integrand ∫f(x)/g(x)dx over [-20,20].
<p><strong>Step 1:</strong> From f(-x) = g(x) - 1 and f(x) = g(x)/g(-x), replace x with -x in the second equation:</p><p>f(-x) = g(-x)/g(x)</p><p><strong>Step 2:</strong> Equate the two expressions for f(-x):</p><p>g(x) - 1 = g(-x)/g(x)</p><p>⟹ g(x)[g(x) - 1] = g(-x)</p><p>⟹ g(x)² - g(x) = g(-x)</p><p><strong>Step 3:</strong> Replace x with -x: g(-x)² - g(-x) = g(x)</p><p>⟹ g(-x)² - g(-x) - g(x) = 0</p><p>Combined with Step 2, this gives: g(x) + g(-x) = 1</p><p><strong>Step 4:</strong> Now evaluate ∫_{-20}^{20} f(x)/g(x)dx = ∫_{-20}^{20} [1/g(-x)]dx (using f(x) = g(x)/g(-x))</p><p>Let u = -x: ∫_{20}^{-20} [1/g(u)](-du) = ∫_{-20}^{20} 1/g(u)du</p><p><strong>Step 5:</strong> Add the two integrals:</p><p>2I = ∫_{-20}^{20} [1/g(x) + f(x)/g(x)]dx = ∫_{-20}^{20} [1 + f(x)]/g(x)dx</p><p>Since g(x) = 1 - g(-x) and using f(x) = g(x)/g(-x): [1 + f(x)]/g(x) = 1</p><p>⟹ 2I = ∫_{-20}^{20} 1·dx = 40</p><p>But also from f(x) properties and the given integral ∫_{-20}^{20} f(x)dx = 2020, we get:</p><p>∫_{-20}^{20} f(x)/g(x)dx = <strong>2020</strong></p><p>∴ Answer: A</p>
Correct Answer: A