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(D) Given the curve $y = x^4 - x^2$. To find the minimum points, we calculate the derivative: $\frac{dy}{dx} = 4x^3 - 2x = 2x(2x^2 - 1) = 0$. This giv

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$\frac{7}{30 \sqrt{2}}$

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(D) Given the curve $y = x^4 - x^2$. To find the minimum points, we calculate the derivative: $\frac{dy}{dx} = 4x^3 - 2x = 2x(2x^2 - 1) = 0$. This gives $x = 0$ and $x = \pm \frac{1}{\sqrt{2}}$. Using the second derivative test, $y'' = 12x^2 - 2$. At $x = \pm \frac{1}{\sqrt{2}}$, $y'' = 12(\frac{1}{2}) - 2 = 4 > 0$, so these are the minimum points. The area is given by $\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} |x^4 - x^2| dx$. Since the function is even, the area is $2 \int_{0}^{\frac{1}{\sqrt{2}}} -(x^4 - x^2) dx$ (as $x^4 - x^2 Area $= 2 \int_{0}^{\frac{1}{\sqrt{2}}} (x^2 - x^4) dx = 2 \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{\frac{1}{\sqrt{2}}}$. $= 2 \left( \frac{1}{3(2\sqrt{2})} - \frac{1}{5(4\sqrt{2})} \right) = 2 \left( \frac{1}{6\sqrt{2}} - \frac{1}{20\sqrt{2}} \right)$. $= 2 \left( \frac{10 - 3}{60\sqrt{2}} \right) = 2 \left( \frac{7}{60\sqrt{2}} \right) = \frac{7}{30\sqrt{2}}$.

The area enclosed (in square units) by the curve $y=x^4-x^2$, the $x$-axis and the vertical lines passing through the two minimum points of the curve is

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