Question

Area bounded by the inverse function of $y=x^3+1$ between the ordinates $x=-7$ and $x=2$ and $x-$axis is

• A. $\frac{51}{4}$
• B. $\frac{26}{4}$
• C. $\frac{51}{2}$
• D. $\frac{26}{2}$

Answer 1

The correct option is A $\frac{51}{4}$


$y=x^3+1=-7\text{or}2\text{(}\because \text{Range of function = domain of inverse function )}$
$\Rightarrow x=-2\text{or}1$
$\therefore$ Required area
$=14+\underset{-2}{\overset{0}{\int }}f\left(x\right)dx+2-\underset{0}{\overset{1}{\int }}f\left(x\right)dx$
$=16+\underset{-2}{\overset{0}{\int }}(x^3+1)dx-\underset{-2}{\overset{0}{\int }}(x^3+1)dx$
$=16-2-\frac{5}{4}=\frac{51}{4}$