If the area enclosed by the parabolas P_1: 2y | vedclass

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(B) The parabolas are $P_1: y = \frac{5x^2}{2}$ and $P_2: y = x^2 + 6$.To find the intersection points,set $\frac{5x^2}{2} = x^2 + 6$,which gives $5x^

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(B) The parabolas are $P_1: y = \frac{5x^2}{2}$ and $P_2: y = x^2 + 6$.To find the intersection points,set $\frac{5x^2}{2} = x^2 + 6$,which gives $5x^2 = 2x^2 + 12$,so $3x^2 = 12$,$x^2 = 4$,$x = \pm 2$.The area $A_1$ enclosed by $P_1$ and $P_2$ is:$A_1 = \int_{-2}^{2} (x^2 + 6 - \frac{5x^2}{2}) dx = 2 \int_{0}^{2} (6 - \frac{3x^2}{2}) dx = 2 [6x - \frac{x^3}{2}]_{0}^{2} = 2(12 - 4) = 16$.The area $A_2$ enclosed by $P_1: y = \frac{5x^2}{2}$ and the line $y = \alpha x$ is found by setting $\frac{5x^2}{2} = \alpha x$,which gives $x = 0$ or $x = \frac{2\alpha}{5}$.$A_2 = \int_{0}^{\frac{2\alpha}{5}} (\alpha x - \frac{5x^2}{2}) dx = [\frac{\alpha x^2}{2} - \frac{5x^3}{6}]_{0}^{\frac{2\alpha}{5}} = \frac{\alpha}{2} (\frac{4\alpha^2}{25}) - \frac{5}{6} (\frac{8\alpha^3}{125}) = \frac{2\alpha^3}{25} - \frac{4\alpha^3}{75} = \frac{6\alpha^3 - 4\alpha^3}{75} = \frac{2\alpha^3}{75}$.Given $A_1 = A_2$,we have $16 = \frac{2\alpha^3}{75}$,so $\alpha^3 = 8 	imes 75 = 600$.

If the area enclosed by the parabolas $P_1: 2y = 5x^2$ and $P_2: x^2 - y + 6 = 0$ is equal to the area enclosed by $P_1$ and the line $y = \alpha x$,where $\alpha > 0$,then $\alpha^3$ is equal to $......$.

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