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Q.
The area between the curve $y=2x^{4}-x^{2}$ , the $x$ -axis and the ordinates of the two minima of the curve is
NTA AbhyasNTA Abhyas 2022Application of Integrals
Solution:
The equation of the curve is $y=2x^{4}-x^{2}=\left(2 x^{2} - 1\right)x^{2}$
The curve is symmetrical about the $y$ -axis.
Also, it is a polynomial of degree four having roots $0,0,\pm\frac{1}{\sqrt{2}}.$
$x=0$ is repeated root. Hence, the graph touches $x$ -axis at $\left(0 , 0\right)$ and intersects the $x$ -axis at
$A\left(- \frac{1}{\sqrt{2}} , 0\right) \, \& \, B\left(\frac{1}{\sqrt{2}} , 0\right)$
$\Rightarrow \frac{d y}{d x}=8x^{3}-2x$
$=2x\left(4 x^{2} - 1\right)=0$
$x=0,x=±\frac{1}{2}$
$\left(\frac{d^{2} y}{d x^{2}}\right)>0$ at $x=\frac{1}{2}$ and at $x=\frac{- 1}{2}$
So, $x=\pm\frac{1}{2}$ are the points of local minima
Thus, the graph of the curve is shown in the diagram.
Here, $y\leq 0$ , as $x$ varies from $x=-\frac{1}{2}$ to $x=\frac{1}{2}$
$\therefore $ The required area
$=2$ Area $OCDO$
$=2\left|\displaystyle \int _{0}^{\frac{1}{2}} y d x\right|$
$=2\left|\displaystyle \int _{0}^{\frac{1}{2}} \left(2 x^{4} - x^{2}\right) d x\right|$
$=\frac{7}{120} \, sq.$ units
