Question

Area bounded by curve $y^3-9y+x=0$ and $y-$axis is

• A. $\frac{9}{2}$
• B. $9$
• C. $\frac{81}{2}$
• D. $81$

Answer 1

The correct option is A $\frac{9}{2}$

Given curve $y^3-9y+x=0,y$-axis i.e., $x=0$

$\Rightarrow x=9y-y^3-\left(i\right)x=0-\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$ we get point of intersection

$9y-y^3=0\Rightarrow y[9-y^2]=0y=0or{y}^2=9\Rightarrow y=\pm 3$

Area bounded is given by

$A={\int }_0^3(9y-y^3)dy={\frac{9y^2}{2}-\frac{y^4}{4}}_0^3=(\frac{9{\left(3\right)}^2}{2}-\frac{{\left(3\right)}^4}{4})-0=\frac{81}{2}-\frac{81}{4}=\frac{81\times 2-81}{4}$

$=\frac{81}{4}$sq units.