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

The two given curves are $$y = x\lvert x\rvert$$ and $$y = x - \lvert x\rvert$$. To integrate, first write each curve without the absolute-value signs.

Case 1: $$x \ge 0$$   • $$\lvert x\rvert = x$$, so $$y_1 = x\lvert x\rvert = x \cdot x = x^{2}$$   • $$y_2 = x - \lvert x\rvert = x - x = 0$$

Case 2: $$x \le 0$$   • $$\lvert x\rvert = -x$$, so $$y_1 = x\lvert x\rvert = x(-x) = -x^{2}$$   • $$y_2 = x - \lvert x\rvert = x - (-x) = 2x$$

Next find the points where the curves intersect.

Case 1 ($$x \ge 0$$): $$x^{2} = 0 \;\Longrightarrow\; x = 0$$

Case 2 ($$x \le 0$$): $$-x^{2} = 2x \;\Longrightarrow\; -x^{2} - 2x = 0 \;\Longrightarrow\; -x(x+2)=0 \;\Longrightarrow\; x = 0 \text{ or } x = -2$$

Thus the two curves meet at $$(-2,-4)$$ and $$(0,0)$$. Between these points the region is completely bounded; outside this interval the graphs do not enclose a finite area.

For $$x \in [-2,0]$$ the upper curve is $$y_1 = -x^{2}$$ and the lower curve is $$y_2 = 2x$$ (check at, say, $$x=-1$$: $$-1^{2}=-1$$ is above $$2(-1)=-2$$).

Required area $$A = \int_{-2}^{0} \bigl[\,(-x^{2}) - (2x)\bigr] \,dx = \int_{-2}^{0} \bigl(-x^{2} - 2x\bigr)\,dx$$

Integrate term by term: $$\int -x^{2}\,dx = -\frac{x^{3}}{3},\qquad \int -2x\,dx = -x^{2}$$

Hence $$A = \Bigl[-\frac{x^{3}}{3} - x^{2}\Bigr]_{-2}^{0}$$

Evaluate the limits: At $$x = 0$$: $$-\frac{0^{3}}{3} - 0^{2} = 0$$
At $$x = -2$$: $$-\dfrac{(-2)^{3}}{3} - (-2)^{2} = -\!\bigl(-\dfrac{8}{3}\bigr) - 4 = \dfrac{8}{3} - 4 = -\dfrac{4}{3}$$

Therefore $$A = 0 - \Bigl(-\dfrac{4}{3}\Bigr) = \dfrac{4}{3}$$

The enclosed area is $$\dfrac{4}{3}$$. This matches Option A.