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(B) Given equations are $y=3x$ and $y=x^2$.To find the points of intersection,we set $3x = x^2$,which gives $x^2 - 3x = 0$,so $x(x-3) = 0$.Thus,the po

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$9/2$

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(B) Given equations are $y=3x$ and $y=x^2$.To find the points of intersection,we set $3x = x^2$,which gives $x^2 - 3x = 0$,so $x(x-3) = 0$.Thus,the points of intersection are $x=0$ and $x=3$.The area of the region bounded by these curves is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=3$.$	ext{Area} = \int_{0}^{3} (3x - x^2) dx$$= \left[ \frac{3x^2}{2} - \frac{x^3}{3} ight]_{0}^{3}$$= \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} ight) - (0 - 0)$$= \left( \frac{27}{2} - \frac{27}{3} ight)$$= \frac{27}{2} - 9$$= \frac{27 - 18}{2} = \frac{9}{2} 	ext{ square units}$.

The area of the region bounded by the line $y=3x$ and the curve $y=x^2$ in square units is

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