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 $y=f\left(x\right)$. If $x\in (-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g\left(x\right)$ satisfying $g\left(0\right)=0$.

The area of the region bounded by the curves $y=f\left(x\right)$ , the $x$-axis, and the lines $x=a$ and $x=b,where-\infty <a<b<-2$, is

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

Answer 1

Answer: (A)

${\int }_a^b\frac{x}{3\left(\right(f\left(x\right){)}^2-1)}dx+bf\left(b\right)-af\left(a\right)$

The required area $={\int }_a^{b}f\left(x\right)dx=xf(x){|}_a^b-{\int }_a^{b}x{f}^{\prime }(x)dx$
$=bf(b)-af(a)+{\int }_a^b\frac{x}{3\left[\right(f(x{)}^2-1)]}dx$.