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Q.
Let $f (x)=x^{3}+3x+2$ and $g (x)$ be the inverse of it. Then he area bounded by $g(x)$, the $x$-axis, and the ordinate at $x = - 2$ and $x = 6$ is
Application of Integrals
Solution:
The required area will be equal to the area enclosed by $y=f (x)$, $y$-axis between the abscissa at $y=-2$ and $y=6$
Hence, $A=\int\limits_{0}^{1} \left(6-f\left(x\right)\right)dx+\int\limits_{-1}^{0}\left(f \left(x\right)-\left(-2\right)\right)dx$
$=\int\limits_{0}^{1}\left(4-x^{3}-3x\right)dx+\int\limits_{-1}^{0}\left(x^{3}+3x+4\right)dx $
$=\frac{5}{4}$ sq units
