The area bounded by the parabola y = 4x^2,the | vedclass

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(C) Given the equation of the parabola is $y = 4x^2$. Expressing $x$ in terms of $y$,we get $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$

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$\frac{7}{3} 	ext{ sq. units}$

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(C) Given the equation of the parabola is $y = 4x^2$. Expressing $x$ in terms of $y$,we get $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$ (since the area is in the first quadrant,we take the positive root). The area bounded by the curve,the $y$-axis,and the lines $y = 1$ and $y = 4$ is given by the integral: $	ext{Area} = \int_{1}^{4} x \, dy = \int_{1}^{4} \frac{\sqrt{y}}{2} \, dy$. $= \frac{1}{2} \int_{1}^{4} y^{1/2} \, dy$. $= \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4} = \frac{1}{2} 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$. $= \frac{1}{3} [4^{3/2} - 1^{3/2}] = \frac{1}{3} [8 - 1] = \frac{7}{3} 	ext{ sq. units}$.

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

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