If the area bounded by the curve x^2=4y,the X | vedclass

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(D) The area $A$ bounded by the curve $y = \frac{x^2}{4}$,the $X$-axis,and the line $x=4$ is given by the integral: $A = \int_{0}^{4} \frac{x^2}{4} dx

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$4^{4/3}$

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(D) The area $A$ bounded by the curve $y = \frac{x^2}{4}$,the $X$-axis,and the line $x=4$ is given by the integral: $A = \int_{0}^{4} \frac{x^2}{4} dx = \left[ \frac{x^3}{12} ight]_{0}^{4} = \frac{64}{12} = \frac{16}{3}$. Since the line $x=\alpha$ divides this area into two equal parts,the area from $x=0$ to $x=\alpha$ must be half of the total area: $\int_{0}^{\alpha} \frac{x^2}{4} dx = \frac{1}{2} 	imes \frac{16}{3} = \frac{8}{3}$. Evaluating the integral: $\left[ \frac{x^3}{12} ight]_{0}^{\alpha} = \frac{\alpha^3}{12} = \frac{8}{3}$. $\alpha^3 = \frac{8 	imes 12}{3} = 8 	imes 4 = 32$. Therefore,$\alpha = (32)^{1/3} = (2^5)^{1/3} = 2^{5/3}$. Re-evaluating the options provided,the correct value is $32^{1/3}$ which corresponds to option $D$.

If the area bounded by the curve $x^2=4y$,the $X$-axis,and the line $x=4$ is divided into two equal areas by the line $x=\alpha$,then the value of $\alpha$ is ...

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