The line y = mx bisects the area enclosed by | vedclass

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(A) The area enclosed by the curve $y = 1 + 4x - x^2$ and the lines $x = 0, x = \frac{3}{2}$ and $y = 0$ is given by the definite integral:$A = \int_{

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$\frac{13}{6}$

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(A) The area enclosed by the curve $y = 1 + 4x - x^2$ and the lines $x = 0, x = \frac{3}{2}$ and $y = 0$ is given by the definite integral:$A = \int_{0}^{3/2} (1 + 4x - x^2) \, dx$$A = \left[ x + 2x^2 - \frac{x^3}{3} ight]_{0}^{3/2}$$A = \left( \frac{3}{2} + 2\left(\frac{9}{4}ight) - \frac{1}{3}\left(\frac{27}{8}ight) ight) - 0$$A = \frac{3}{2} + \frac{9}{2} - \frac{9}{8} = 6 - \frac{9}{8} = \frac{48 - 9}{8} = \frac{39}{8}$The line $y = mx$ bisects this area,so the area of the triangle formed by the line $y = mx$,the $x$-axis,and the line $x = \frac{3}{2}$ must be half of the total area.The triangle has vertices at $(0,0)$,$(\frac{3}{2}, 0)$,and $(\frac{3}{2}, \frac{3m}{2})$.Area of triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{3}{2} 	imes \frac{3m}{2} = \frac{9m}{8}$Equating the triangle area to half the total area:$\frac{9m}{8} = \frac{1}{2} 	imes \frac{39}{8}$$9m = \frac{39}{2}$$m = \frac{39}{18} = \frac{13}{6}$

The line $y = mx$ bisects the area enclosed by the curve $y = 1 + 4x - x^2$ and the lines $x = 0, x = \frac{3}{2}$ and $y = 0$. Then the value of $m$ is:

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