Let A be the area bounded by the curve y=x|x- | vedclass

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(C) The function is given by $y = x|x-3|$. For the interval $[-1, 2]$,$x-3$ is always negative,so $|x-3| = -(x-3) = 3-x$.Thus,$y = x(3-x) = 3x - x^2$.

Vedclass · Accepted Answer

$62$

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(C) The function is given by $y = x|x-3|$. For the interval $[-1, 2]$,$x-3$ is always negative,so $|x-3| = -(x-3) = 3-x$.Thus,$y = x(3-x) = 3x - x^2$.Since $3x - x^2$ is positive for $x \in [0, 3]$,the area $A$ is given by the integral:$A = \int_{-1}^{2} (3x - x^2) dx$$A = \left[ \frac{3x^2}{2} - \frac{x^3}{3} ight]_{-1}^{2}$$A = \left( \frac{3(4)}{2} - \frac{8}{3} ight) - \left( \frac{3(1)}{2} - \frac{-1}{3} ight)$$A = \left( 6 - \frac{8}{3} ight) - \left( \frac{3}{2} + \frac{1}{3} ight)$$A = \left( \frac{18-8}{3} ight) - \left( \frac{9+2}{6} ight) = \frac{10}{3} - \frac{11}{6} = \frac{20-11}{6} = \frac{9}{6} = \frac{3}{2}$.Wait,let us re-evaluate the integral:$A = \int_{-1}^{2} (3x - x^2) dx = \left[ \frac{3x^2}{2} - \frac{x^3}{3} ight]_{-1}^{2} = (6 - 8/3) - (3/2 + 1/3) = 10/3 - 11/6 = 9/6 = 1.5$.Re-checking the curve: $y = x|x-3|$. For $x \in [-1, 0]$,$x$ is negative and $(x-3)$ is negative,so $y = x(3-x) = 3x - x^2$. For $x \in [0, 2]$,$x$ is positive and $(x-3)$ is negative,so $y = x(3-x) = 3x - x^2$.Since $3x - x^2$ is negative for $x \in [-1, 0]$ and positive for $x \in [0, 2]$,the area is:$A = \int_{-1}^{0} -(3x - x^2) dx + \int_{0}^{2} (3x - x^2) dx$$A = \int_{-1}^{0} (x^2 - 3x) dx + \int_{0}^{2} (3x - x^2) dx$$A = [x^3/3 - 3x^2/2]_{-1}^{0} + [3x^2/2 - x^3/3]_{0}^{2}$$A = (0 - (-1/3 - 3/2)) + (6 - 8/3 - 0) = (1/3 + 3/2) + (10/3) = 11/6 + 20/6 = 31/6$.Therefore,$12A = 12 	imes (31/6) = 62$.

Let $A$ be the area bounded by the curve $y=x|x-3|$,the $x$-axis and the ordinates $x=-1$ and $x=2$. Then $12A$ is equal to $...........$.

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