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(D) To find the area of the region bounded by the curves $y=9x^2$ and $y=5x^2+4$,we first find their points of intersection by setting the equations e

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$\frac{16}{3}$

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(D) To find the area of the region bounded by the curves $y=9x^2$ and $y=5x^2+4$,we first find their points of intersection by setting the equations equal to each other:$9x^2 = 5x^2 + 4$$4x^2 = 4$$x^2 = 1$$x = \pm 1$When $x = 1$,$y = 9(1)^2 = 9$. When $x = -1$,$y = 9(-1)^2 = 9$.So,the points of intersection are $(1, 9)$ and $(-1, 9)$.The required area is the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$:$	ext{Area} = \int_{-1}^{1} [(5x^2+4) - 9x^2] dx$Since the region is symmetric about the $y$-axis,we can write:$	ext{Area} = 2 \int_{0}^{1} (4 - 4x^2) dx$$= 2 [4x - \frac{4x^3}{3}]_{0}^{1}$$= 2 (4(1) - \frac{4(1)^3}{3}) - 2(0 - 0)$$= 2 (4 - \frac{4}{3})$$= 2 (\frac{12-4}{3})$$= 2 (\frac{8}{3}) = \frac{16}{3} 	ext{ sq units}$.

The area of the region bounded by the curves $y=9x^2$ and $y=5x^2+4$ (in sq units) is

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