Question

Consider the function \(y=f(x)\) defined implicitly by the equation \[ y^{3}-3y+x=0 \] on the interval \((-\infty,-2)\cup(2,\infty)\). The area of the region bounded by the curve \(y=f(x)\), the x-axis and the lines \(x=a,x=b\), where \(-\infty<a<b<-2\) is:

• A. $\int_{a}^{b}\frac{x\,dx}{3((f(x))^{2}-1)}-b\,f(b)+a\,f(a)$
• B. $-\int_{a}^{b}\frac{x\,dx}{3((f(x))^{2}-1)}-b\,f(b)+a\,f(a)$
• C. $\int_{a}^{b}\frac{x\,dx}{3((f(x))^{2}-1)}+b\,f(b)-a\,f(a)$
• D. $-\int_{a}^{b}\frac{x\,dx}{3((f(x))^{2}-1)}+b\,f(b)-a\,f(a)$

Hint:

Whenever the curve lies below the x-axis, remember that area is computed using $$ \text{Area}=-\int y\,dx $$ instead of $\int y\,dx$.

Answer 1

The Correct Option is A

Concept: The area bounded by a curve below the x-axis is: $$ \text{Area}=-\int_a^b y\,dx $$ For implicit functions, integration by parts is useful: $$ \int y\,dx = xy-\int x\,dy $$ Step 1: Express $x$ in terms of $y$.
Given: $$ y^3-3y+x=0 $$ Therefore, $$ x=3y-y^3 $$ Differentiate with respect to $y$: $$ \frac{dx}{dy}=3-3y^2=-3(y^2-1) $$ Hence, $$ dx=-3(y^2-1)\,dy $$ So, $$ dy=-\frac{dx}{3(y^2-1)} $$ Since $y=f(x)$, $$ dy=-\frac{dx}{3((f(x))^2-1)} $$

Step 2: Set up the area integral.
For $x\in(-\infty,-2)$, the curve lies below the x-axis, so: $$ \text{Area}=-\int_a^b y\,dx $$ Using integration by parts: $$ \int_a^b y\,dx = \Big[xy\Big]_a^b-\int_{f(a)}^{f(b)}x\,dy $$ Thus, $$ \text{Area} = -\left( b\,f(b)-a\,f(a)-\int_{f(a)}^{f(b)}x\,dy \right) $$ Simplifying: $$ \text{Area} = -b\,f(b)+a\,f(a)+\int_{f(a)}^{f(b)}x\,dy $$

Step 3: Convert the integral into $x$-form.
Using $$ dy=-\frac{dx}{3((f(x))^2-1)} $$ we get: $$ \int_{f(a)}^{f(b)}x\,dy = \int_a^b x\left( -\frac{dx}{3((f(x))^2-1)} \right) $$ Therefore, $$ \int_{f(a)}^{f(b)}x\,dy = -\int_a^b \frac{x\,dx}{3((f(x))^2-1)} $$ Substituting back: $$ \text{Area} = -b\,f(b)+a\,f(a) + \int_a^b \frac{x\,dx}{3((f(x))^2-1)} $$ Hence, $$ \boxed{ \text{Area} = \int_a^b \frac{x\,dx}{3((f(x))^2-1)} -b\,f(b)+a\,f(a) } $$ This matches option (A).