# Area with a Twist(Corrected)

**Calculus**Level 5

If the Area bounded by \(y=f(x)\) ,\(x=\dfrac{1}{2}\),\(x=\dfrac{\sqrt{3}}{2} \) and the x-axis, is \(A\) sq. units, where \[f(x) = x+ \dfrac{2}{3}x^3 + \dfrac{2}{3}\cdot\dfrac{4}{5}x^5 + \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}x^7+ \cdots \infty \] and \(|x| < 1\) and \(A\) can be written as \[ \dfrac{\pi^a}{b} \] where \(a,b\) are positive integers. Find \( a + b \).