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

###### For more problems on calculus, click here.

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

Already have an account? Log in here.