MCQ
Area bounded by curve $x(x^2 + p)$ = $y -1$ with $y = 1$ is
- A$\frac{{{p^2}}}{4}$
- B$\frac{p}{2}$
- ✓$\frac{{{p^2}}}{2}$
- D$\frac{p}{4}$
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} $
$-\left(\frac{\lambda^{4}}{4}-\frac{\lambda^{4}}{2}-\lambda\right)-\lambda+\left(\lambda-\left[\frac{\lambda^{4}}{4}-\frac{\lambda^{2}}{2}+\lambda\right]\right)$
$\left(\frac{\lambda^{4}}{4}\right)+\left(\frac{\lambda^{4}}{4}\right)=\frac{\lambda^{4}}{2}=\frac{\mathrm{p}^{2}}{2}$
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$ \frac{\pi}{2}$