The area between the curve y = 2x^ 4 - x^ 2 , the x -axis and the...

The equation of the curve is $y = 2x^{4} - x^{2} = \left(2x^{2} - 1\right)x^{2}$

The curve is symmetrical about the $\text{y}$-axis.

Also, it is a polynomial of degree four having roots $0, 0, \pm \dfrac{1}{\sqrt{2}} .<br /><br />\text{x} =$0$is repeated root. Hence, graph touches \text{x} -axis at \text{(0,0)}$ and intersects the \text{x} -axis at \text{A} \left(- \dfrac{1}{\sqrt{2}} ,0\right) & \text{B}\left(\dfrac{1}{\sqrt{2}} ,0\right)<br /><br />⇒ $\dfrac{\text{dy}}{\text{dx}} =8x^{3} - 2 \text{x}<br /><br />$= 2 $\text{x}$ (4x^{2} - 1 ) = 0$<br /><br />\text{x} =0, \text{x} =\pm \dfrac{1}{2}<br /><br /> \left(\dfrac{d^{2} \text{y}}{\text{d}x^{2}}\right)$> 0$at \text{x} =\dfrac{1}{2}$ and at $\text{x}=\dfrac{- 1}{2}$

So x = \pm $\dfrac{1}{2} are$ the points of local minima

Thus, the graph of the curve is shown in diagram.



Here,y ≤ 0$, as \text{x} varies from$ x = - $\dfrac{1}{2}$ to x =$\dfrac{1}{2}<br /><br />$∴$The required area<br /><br />$= 2AreaOCDO$<br /><br />$= 2\left|\mathop{∫_{{0}}$^{\dfrac{1}{2}}$ydx\right|$<br /><br />$= 2\left|\mathop{∫_{{0}}$^{\dfrac{1}{2} }\left(2x^{4} - x^{2}\right)$dx\right|$<br /><br />=\dfrac{7}{120}$sq . units