I don't think I can do it 4

Without loss of generality, swap variable $x$ with $y$ , then we have: $\begin{cases} y = x^2-2x \\ y = - x \end{cases}$ .

We note that the area is bounded within $0 \le x \le 1$ (see graph). Therefore, the area is given by:

$\begin{aligned} A & = \left|\int_0^1 (x^2-2x) \space dx - \int_0^1 (-x) \space dx \right| \\ & = \left|\left[\frac{x^3}{3}-x^2\right]^1_0 - \left[-\frac{x^2}{2}\right]^1_0 \right| \\ & = \left| - \frac{2}{3} + \frac{1}{2} \right| = \left| - \frac{1}{6} \right| = \frac{1}{6} \end{aligned}$

$\Rightarrow A = \boxed{6}$

without swapping variables: $$ for the required area, the bounds are -1,0. and the equation $x=y^2-2y\Longrightarrow y^2-2y+1=x+1\Longrightarrow y=\pm\sqrt{x+1}+1$ obviously for the required range $y=-\sqrt{x+1}+1$ . $$ now we just integrate: $\left|\int_{-1}^0 \left(-\sqrt{x+1}+1\right)dx-\int_{-1}^0 \left(-x\right)dx\right|$ $\left|\left[-\dfrac{(x+1)^{3/2}}{\dfrac{3}{2}}+x\right]_{-1}^0-\left[-\dfrac{x^2}{2}\right]_{-1}^0\right|$ $\left|\dfrac{1}{3}-\left(\dfrac{1}{2}\right)\right|=\dfrac{1}{\boxed{6}}$