Question

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 $g(0)=0$
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

• A. $\underset{a}{\overset{b}{\int }}\frac{x}{3\left((f(x){)}^2-1\right)}dx+bf(b)-af(a)$
• B. $-\underset{a}{\overset{b}{\int }}\frac{x}{3\left((f(x){)}^2-1\right)}dx+bf(b)-af(a)$
• C. $\underset{a}{\overset{b}{\int }}\frac{x}{3\left((f(x){)}^2-1\right)}dx-bf(b)+af(a)$
• D. $-\underset{a}{\overset{b}{\int }}\frac{x}{3\left((f(x){)}^2-1\right)}dx-bf(b)+af(a)$

Answer 1

Answer: (A)

$=-af(a)-(-bf(b))-\underset{f(b)}{\overset{f(a)}{\int }}-\left(3y-y^3\right)dy$
$=bf(b)-af(a)+\underset{b}{\overset{a}{\int }}x{f}^{\prime }(x)dx$, putting $y=f(x)$
$=bf(b)-af(a)+\underset{b}{\overset{a}{\int }}\frac{x}{3\left(1-y^2\right)}dx\left(\because \frac{dx}{dy}=3-3y^2\equiv \frac{1}{f^{\prime }(x)}\Rightarrow f^{\prime }(x)=\frac{1}{3\left(1-(f(x){)}^2\right)}\right)$
$=bf(b)-af(a)-\underset{a}{\overset{b}{\int }}\frac{x}{3\left(1-(f(x){)}^2\right)}dx=bf(b)-af(a)+\underset{a}{\overset{b}{\int }}\frac{x}{3\left((f(x){)}^2-1\right)}dx$
(1) is correct option
Aliter $\because y^3-3y+x=0$
So $y^{\prime }=\frac{-1}{3\left(y^2-1\right)}$
$bf(b)-af(a)=\mid x{f(x)|}_a^b=\underset{a}{\overset{b}{\int }}\left\{x{f}^{\prime }(x)+f(x)\right\}dx$
So area $\Delta =\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}\left(f(x)+x{f}^{\prime }(x)-x{f}^{\prime }(x)\right)dx$
$=xf(x){|}_a^b-|xf(x){|}_a^b$
$=bf(b)-af(a)+{\int }_a^b\frac{x}{3\left(f(x{)}^2-1\right)}dx$