Let's do some calculus! (16)

Consider the functions defined implicitly by the equation $y^{3} - 3y + x = 0$ on various intervals in the real line. If $x \in (-\infty,-2) \cup (2, \infty)$ , the equation implicitly defines a unique real valued differentiable function $y = f(x)$ . If $x \in (-2,2)$ , the equation implicitly defines a unique real valued differentiable function $y = g(x)$ satisfying $g(0) = 0$ . What is the area of the region bounded by the curves $y = f(x)$ , and the lines $y=0$ , $x=a$ and $x=b$ , where $-\infty < a < b < -2$ ? Select your answer from the given options.

A) $\displaystyle \int_{a}^{b}{\dfrac{x}{3\big( {\left( f(x) \right)}^{2} - 1 \big)}} \,dx + b f(b) - a f(a)$

B) $\displaystyle -\int_{a}^{b}{\dfrac{x}{3\big( {\left( f(x) \right)}^{2} - 1 \big)}} \,dx + b f(b) - a f(a)$

C) $\displaystyle \int_{a}^{b}{\dfrac{x}{3\big( {\left( f(x) \right)}^{2} - 1 \big)}} \,dx - b f(b) + a f(a)$

D) $\displaystyle -\int_{a}^{b}{\dfrac{x}{3\big( {\left( f(x) \right)}^{2} - 1 \big)}} \,dx - b f(b) + a f(a)$

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