Question

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

• A. $\frac{7}{120}sq.units$
• B. $\frac{11}{120}sq.units$
• C. $\frac{13}{120}sq.units$
• D. $\frac{8\sqrt{2}-7}{120}sq.units$

Answer 1

Answer: (A)

$\frac{7}{120}sq.units$

$y=2x^4-x^2\Rightarrow \frac{dy}{dx}=8x^3-2x=2x(4x^2-1)=0$, for extrema.
$\Rightarrow x=\pm \frac{1}{2}$
Thus required area $=-2{\int }_0^{1/2}(2x^4-x^2)dx=\frac{7}{120}$