For which of the following values of m,the ar | vedclass

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(B) The equation of the curve is $y = x - x^2$.The points of intersection of the curve $y = x - x^2$ and the line $y = mx$ are found by setting $x - x

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$-2$

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(B) The equation of the curve is $y = x - x^2$.The points of intersection of the curve $y = x - x^2$ and the line $y = mx$ are found by setting $x - x^2 = mx$,which gives $x(1 - x - m) = 0$. Thus,$x = 0$ or $x = 1 - m$.The area $A$ of the region bounded by the curve and the line is given by the integral:$A = \int_{0}^{1-m} (x - x^2 - mx) dx = \int_{0}^{1-m} ((1 - m)x - x^2) dx$Evaluating the integral:$A = \left[ (1 - m)\frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1-m}$$A = (1 - m)\frac{(1 - m)^2}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{6}$Given that the area is $\frac{9}{2}$,we have:$\frac{(1 - m)^3}{6} = \frac{9}{2}$$(1 - m)^3 = 27$$1 - m = 3$$m = -2$Thus,the correct value of $m$ is $-2$.

For which of the following values of $m$,the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $\frac{9}{2}$?

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