The area enclosed between the curves y=x|x| a | vedclass

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(D) First,we define the curves for different intervals of $x$:For $x \ge 0$,$y = x(x) = x^2$ and $y = x - x = 0$.For $x < 0$,$y = x(-x) = -x^2$ and $y

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$\frac{4}{3}$

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(D) First,we define the curves for different intervals of $x$:For $x \ge 0$,$y = x(x) = x^2$ and $y = x - x = 0$.For $x To find the intersection points for $x The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -2$ to $x = 0$:$A = \int_{-2}^{0} (2x - (-x^2)) \, dx = \int_{-2}^{0} (x^2 + 2x) \, dx$$A = \left[ \frac{x^3}{3} + x^2 \right]_{-2}^{0}$$A = (0 + 0) - \left( \frac{(-2)^3}{3} + (-2)^2 \right) = - \left( -\frac{8}{3} + 4 \right) = - \left( \frac{4}{3} \right) = -\frac{4}{3}$.Since area must be positive,we take the absolute value: $|-\frac{4}{3}| = \frac{4}{3}$.

The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

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