question_answer

Let\[f(x)={{x}^{3}}+3x+2\] and g(x) be the inverse of it. Then the area bounded by g(x), the x-axis, and the ordinate at \[x=-\,2\] and \[x=6\] is

A) 1/4 sq. units      

B) 4/3 sq. units

C) 5/4 sq. units      

D) 7/3 sq. units

Correct Answer: C

Solution :

[c] 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, \[A=\int\limits_{0}^{1}{\left( 6-f(x) \right)}\,dx+\int\limits_{-1}^{0}{\left( f(x)-(-2) \right)}\,dx\] \[=\int\limits_{0}^{1}{\left( 4-{{x}^{3}}-3x \right)dx+\int\limits_{-1}^{0}{\left( {{x}^{3}}+3x+4 \right)}dx=\frac{5}{4}}\]Sq. units.