<p>If <br>
\[ f(x) = x + \frac{2}{3}x^3 + \frac{2}{3}\cdot\frac{4}{5}x^5 + \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7 + \cdots \infty \]
then the area \(A = \int_{1/2}^{\sqrt{3}/2} \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx = \frac{\pi^2}{24}\). If \(A = \frac{\pi^2}{a+b}\) where \(a+b\) equals:</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is the Taylor series of sinh⁻¹(x) or arcsin(x) related function, and use substitution u = sin⁻¹(x) to convert the integral into a standard form that evaluates to π²/24, revealing the connection between the series and the definite integral.
<p><strong>Step 1:</strong> Use substitution u = sin⁻¹(x), then du = dx/√(1-x²)</p><p><strong>Step 2:</strong> When x = 1/2, u = π/6; when x = √3/2, u = π/3</p><p><strong>Step 3:</strong> The integral becomes A = ∫_{π/6}^{π/3} u du = [u²/2]_{π/6}^{π/3}</p><p><strong>Step 4:</strong> A = (π/3)²/2 - (π/6)²/2 = π²/18 - π²/72 = (4π² - π²)/72 = 3π²/72 = π²/24</p><p><strong>Step 5:</strong> Given A = π²/(a+b) = π²/24, therefore a + b = <strong>24</strong></p>
Correct Answer: 26