;)
Son dos recintos de igual área:
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calculo puntos de corte:
$$\begin{align}&y=2x^3-1\\&y=x-1\\&2x^3-1=x-1\\&2x^3-x=0\\&x(2x^2-1)=0\\&x=0\\&x= \pm \frac{1}{\sqrt 2}\\&\\&Area=2 \int_0^\frac{1}{\sqrt 2}x-1-(2x^2-1)dx=\\&\\&2 \int_0^\frac{1}{\sqrt 2} -2x^3+x)dx=\\&\\&2 \Bigg[ - \frac{x^4}{2}+ \frac{x^2}{2} \Bigg]_0^\frac{1}{\sqrt 2}= -x^4 + x^2 \Bigg|_0^\frac{1}{\sqrt 2}=\\&\\&\frac{-1}{4}+ \frac{1}{2}= \frac{1}{4} \ \ u^2\end{align}$$
saludos
;)
;)