Question

The area bounded by the curve $y=2x^4-x^2$, x-axis and the two ordinates corresponding to the minima of the function is $\frac{8}{a}$. Find $a$.

Answer 1

$y=2x^4-x^2$

$\frac{dy}{dx}=8x^3-2x$

$=2x(4x^2-1)$

$x=0$ $x=\pm \frac{1}{2}$

Hence

$Area={\int }_{-1/2}^{1/2}2x^4-x^{2}dx$

As even function$=2{\int }_0^{1/2}2x^4-x^{2}dx$

$={[\frac{2x^5}{5}-\frac{x^3}{3}]}_0^1$

$=\frac{8}{120}$ As the curve is below $X-$axis.

Area$=\frac{8}{120}$