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The area bounded by the curve $4y^2 = x^2(4-x)(x-2)$ is equal to:
(1) $\frac{\pi}{8}$
(2) $\frac{3\pi}{8}$
(3) $\frac{3\pi}{2}$
(4) $\frac{\pi}{16}$
Domain $x \in [2,4]$. Using the property $\int_a^b f(x),dx = \int_a^b f(a+b-x),dx$: $A = 3 \cdot \frac{\pi}{2} \cdot 1^2 = \frac{3\pi}{2}$.
BETA
