Explanation

(a)
$\because y = 2x^{4} - x^{2}$
$\therefore\frac{dy}{dx} = 8x^{3} - 2x$
For maxima or minima, put $\frac{dy}{dx} = 0$, we get
$x = - \frac{1}{2},0,\frac{1}{2}$
Then, $\left( \frac{d^{2}y}{dx^{2}} \right)_{x = \frac{1}{2}} > 0,\ \left( \frac{d^{2}y}{dx^{2}} \right)_{x = 0} < 0$
and $\left( \frac{d^{2}y}{dx^{2}} \right)_{x = \frac{1}{2}} > 0$
$\therefore$ Required area $= \left| \int_{- 1/2}^{1/2}{\left( 2x^{4} - x^{2} \right)dx} \right| = \frac{7}{120}\ sq\ unit$