Question

The area between the curve $y=2x^4-x^2,$ the $x-$axis and the ordinates of two minima of the curve is:

• A. $\frac{7}{120}$ sq.unit
• B. $\frac{9}{120}$ sq.unit
• C. $\frac{11}{120}$ sq.unit
• D. $\frac{13}{120}$ sq.unit

Answer 1

The correct option is A $\frac{7}{120}$ sq.unit

$y=2x^4-x^2$
$\therefore \frac{dy}{dx}=8x^3-2x,$ for max. or min. $\frac{dy}{dx}=0$
$x=-\frac{1}{2},0,\frac{1}{2}$
Then, ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-\frac{1}{2}}>0,{\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}<0$
and ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{1}{2}}>0$
$\therefore$Required Area$=\left|{\int }_{-\frac{1}{2}}^{\frac{1}{2}}(2x^4-x^2)dx\right|$
$={[\frac{2x^5}{5}-\frac{x^3}{3}]}_{-\frac{1}{2}}^{\frac{1}{2}}$
$=\frac{7}{120}$.sq.unit.