Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?

By differentiation under the integral sign we get
\begin{align*}\left(\int_0^y\sqrt{x^4+(y-y^2)^2}\,dx\right)'&=\int_0^y\frac{2(y-y^2)(1-2y)}{2\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}\\&=\sqrt{y^4+(y-y^2)^2}+\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx,\end{align*}
which is positive for $y\in(0,\frac12)$. Else suppose $y\in(\frac12,1)$, then it's positive if
$$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx=-\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx<\sqrt{y^4+(y-y^2)^2}=y\sqrt{y^2+(1-y)^2}.$$
Clearly,
$$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx<y\cdot\frac{(2y-1)(y-y^2)}{\sqrt{(y-y^2)^2}}=y(2y-1),$$
so it remains to check that
$$2y-1\le\sqrt{y^2+(1-y)^2}\iff 0\le y^2+(1-y)^2-(2y-1)^2=-2y^2+2y=2y(1-y)$$
So we can conclude the maximum is attained for $y=1$, where it's
$$\int_0^1\sqrt{x^4}\,dx=\int_0^1 x^2\, dx=\frac13.$$
