Paragraph: Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in…
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ 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 $g(0)=0$.Question:
The area of the region bounded by the curve $y=f(x)$, the $X$-axis and the lines $x=a$ and $x=b$, where $-\infty < a < b < -2$, is
$\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)$
$-\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x+b f(b)-a f(a)$
$\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x-b f(b)+a f(a)$
$-\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x-b f(b)-a f(a)$
Solution
Required area $=\int_a^b y d x=\int_a^b f(x) d x$
$
=[f(x) \cdot x]_a^b-\int_a^b f^{\prime}(x) \cdot x d x=b f(b)-a f(a)-\int_a^b f^{\prime}(x) \cdot x d x
$
$
=b f(b)-a f(a)+\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x
$
As, $\quad f^{\prime}(x)=\frac{d y}{d x}=-\frac{1}{3\left(y^2-1\right)}=-\frac{1}{3\left[(f(x))^2-1\right]}$