Cho ${{x}^{2}}+x-1={{x}^{4}}+x-1\Leftrightarrow {{x}^{2}}-{{x}^{4}}=0$

$\Leftrightarrow x=0\vee x=\pm 1$

Khi đó diện tích cần tìm là $S=\int_{-1}^{1}{\left| {{x}^{2}}-{{x}^{4}} \right|dx}$, hay $S=\left| \int_{-1}^{0}{\left( {{x}^{2}}-{{x}^{4}} \right)dx} \right|+\left| \int_{0}^{1}{\left( {{x}^{2}}-{{x}^{4}} \right)dx} \right|$

$=\left| \left. \left( \frac{{{x}^{3}}}{3}-\frac{{{x}^{5}}}{5} \right) \right|_{-1}^{0} \right|+\left| \left. \left( \frac{{{x}^{3}}}{3}-\frac{{{x}^{5}}}{5} \right) \right|_{0}^{1} \right|=\frac{4}{15}$