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:
$\int_{-1}^1 g^{\prime}(x) d x$ is equal to
$2 g(-1)$
0
$-2 g(1)$
$2 g(1)$
Solution
Let $I=\int_{-1}^1 g^{\prime}(x) d x=[g(x)]_{-1}^1=g(1)-g(-1)$
Since, $\quad y^3-3 y+x=0$ and $\quad y=g(x)$
$
\therefore(g(x))^3-3 g(x)+x=0
$
[by Eq. (i)]
At $\quad x=1$,
$
\begin{aligned}
(g(1))^3-3 g(1)+1 & =0 \\
\text { At } \quad x & =-1, \\
(g(-1))^3-3 g(-1)-1 & =0
\end{aligned}
$
On adding Eqs. (i) and (ii), we get
$
\begin{array}{rlrl}
& & (g(1))^3+(g(-1))^3-3(g(1)+g(-1))=0 \\
\Rightarrow & & {[g(1)+g(-1)]\left[(g(1))^2+(g(-1))^2-g(1) g(-1)-3\right]=0} \\
\Rightarrow & g(1)+g(-1)=0 \\
\Rightarrow & g(1) & =-g(-1) \\
& \therefore & I & =g(1)-g(-1)=g(1)-(-g(1))=2 g(1)
\end{array}
$