# 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 $$a + b$$.

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