An application of Green's theorem

Intersection. $x^2=x$ gives $x=0$ or $1$.

Lower path. $\gamma_1(t)=(t,t)$, $t\in[0,1]$. Contribution $\frac12\int_0^1(t\cdot1-t\cdot1)\,dt=0$.

Upper path. $\gamma_2(t)=(t,t^2)$ from $t=1$ to $0$ (reverse). Compute $\frac12\int_{1}^{0}(t\cdot2t-t^2\cdot1)\,dt=\frac12\int_1^0 t^2\,dt=\frac12\cdot(-\frac13)=-\frac16$; absolute value $\frac16$.

Wait, sign issue: total area should be positive. Let me just compute $|x^2-x|$ directly: area $=\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16$.

Conclusion. Area $=1/6$.