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(D) The given parabola is $y = 9x^2$,which implies $x^2 = y/9$. Since we are in the first quadrant,$x = \sqrt{y}/3$.The area $A$ bounded by the curve

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$14/9$

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(D) The given parabola is $y = 9x^2$,which implies $x^2 = y/9$. Since we are in the first quadrant,$x = \sqrt{y}/3$.The area $A$ bounded by the curve $x = f(y)$,the $y$-axis $(x = 0)$,and the lines $y = 1$ and $y = 4$ is given by the integral:$A = \int_{1}^{4} x \, dy = \int_{1}^{4} \frac{\sqrt{y}}{3} \, dy$$A = \frac{1}{3} \int_{1}^{4} y^{1/2} \, dy$Using the power rule $\int y^n \, dy = \frac{y^{n+1}}{n+1}$,we get:$A = \frac{1}{3} \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4} = \frac{1}{3} 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$$A = \frac{2}{9} \left( 4^{3/2} - 1^{3/2} ight)$$A = \frac{2}{9} (8 - 1) = \frac{2}{9} 	imes 7 = \frac{14}{9} 	ext{ sq. units.}$

The area of the region (in sq. units),in the first quadrant bounded by the parabola $y = 9x^2$ and the lines $x = 0, y = 1$ and $y = 4$,is

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