Area enclosed by curves

Question

Given the curves $y=x^2$ and $y=\frac{1}{2}(x+x^4)$. What is the area enclosed by them ? I can't find the points of intersection of the curves.

Answer 1

Since $$\begin{align}x^2\ge \frac{x+x^4}{2}&\iff x^4-2x^2+x\le 0\\&\iff x(x-1)\left(x-\frac{-1+\sqrt 5}{2}\right)\left(x-\frac{-1-\sqrt 5}{2}\right)\le 0 \\&\iff \frac{-1-\sqrt 5}{2}\le x\le 0\ \ \text{or}\ \ \frac{-1+\sqrt 5}{2}\le x\le 1,\end{align}$$ the area is $$\int_{\frac{-1-\sqrt 5}{2}}^{0}\left(x^2-\frac{x+x^4}{2}\right)dx+\int_{0}^{\frac{-1+\sqrt 5}{2}}\left(\frac{x+x^4}{2}-x^2\right)dx+\int_{\frac{-1+\sqrt 5}{2}}^{1}\left(x^2-\frac{x+x^4}{2}\right)dx.$$ Here, use $x^2=-x+1$ for $x=\frac{-1\pm\sqrt 5}{2}$.