STD 12 - 8. Application and integration — Mathematics STD 12 Science — Question

say $\mathrm{p}=-\lambda^{2}$

$y=x^{3}-\lambda^{2} x+1$

point of intersection with $y=1$

$\mathrm{x}=0, \mathrm{x}=\lambda, \mathrm{x}=-\lambda$

$\left[ {\int\limits_{ - 2}^0 {\left( {{x^3} - {\lambda ^2}x + 1} \right)dx - \lambda } } \right] + \lambda  - \int\limits_0^\lambda  {\left( {{x^3} - \lambda {x^2} + 1} \right)dx} $

$-\left(\frac{\lambda^{4}}{4}-\frac{\lambda^{4}}{2}-\lambda\right)-\lambda+\left(\lambda-\left[\frac{\lambda^{4}}{4}-\frac{\lambda^{2}}{2}+\lambda\right]\right)$

$\left(\frac{\lambda^{4}}{4}\right)+\left(\frac{\lambda^{4}}{4}\right)=\frac{\lambda^{4}}{2}=\frac{\mathrm{p}^{2}}{2}$