Question

# Find the area of the region bounded by the graphs of the given functions y=x⁴−4x² y=4x² Give your numerical answer $approximate number only$ in square units to three decimal places of precision.

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## Answer to a math question Find the area of the region bounded by the graphs of the given functions y=x⁴−4x² y=4x² Give your numerical answer $approximate number only$ in square units to three decimal places of precision.

Frederik
4.6
64 Answers
To find the area of the region bounded by the graphs of the functions y = x^4 - 4x^2 and y = 4x^2 , we first need to find the points of intersection:

Setting the two equations equal to each other, we have:
x^4 - 4x^2 = 4x^2
x^4 - 8x^2 = 0
x^2$x^2 - 8$ = 0
This gives us x = 0, \sqrt{8}, -\sqrt{8} as points of intersection.

The area of the region bounded by the two functions is given by:
A = 2 \times \int_{0}^{\sqrt{8}} |4x^2 - $x^4 - 4x^2$| dx

Simplify the absolute value expression inside the integral:
A = 2 \times \int_{0}^{\sqrt{8}} |4x^2 - x^4 + 4x^2| dx
A = 2 \times \int_{0}^{\sqrt{8}} |8x^2 - x^4| dx

Further simplifying:
A = 2 \times \int_{0}^{\sqrt{8}} $x^4 - 8x^2$ dx

Integrating term by term:
A = 2 \times $\frac{x^5}{5} - \frac{8x^3}{3}$ \Big|_{0}^{\sqrt{8}}
A = 2 \times $\frac{8\sqrt{8}}{5} - \frac{8 \cdot 8\sqrt{8}}{3}$
A = \frac{16\sqrt{8}}{5} - \frac{128\sqrt{8}}{3}
A = \frac{48\sqrt{8}}{15}

Therefore, the area of the region bounded by the graphs of the functions y = x^4 - 4x^2 and y = 4x^2 is \frac{48\sqrt{8}}{15} square units.

\boxed{A = \frac{48\sqrt{8}}{15}}=48.272

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