# Let's do some calculus! (16)

Calculus Level 4

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