Question

Area bounded by the curve $x(x^2+P)=y-1$ with $y=1$ is

• A. $\frac{P^2}{4}$
• B. $\frac{P}{2}$
• C. $\frac{P^2}{2}$
• D. $\frac{P}{4}$

Answer 1

The correct option is C $\frac{P^2}{2}$

$y=x^3+Px+1$

Let $P=-{\lambda }^2\Rightarrow y=x^3-{\lambda }^{2}x+1,$
Point of intersection with $y=1.$
$x=0,x=\lambda ,x=-\lambda$
Therefore,
$(\underset{-\lambda }{\overset{0}{\int }}(x^3-{\lambda }^{2}x+1)dx-\lambda )+\lambda -(\underset{0}{\overset{\lambda }{\int }}(x^3-{\lambda }^{2}x+1)dx)=\frac{{\lambda }^4}{2}=\frac{P^2}{2}$