Find the maximum value Find the maximum value $$F(y)=\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$$ with $y\in [0;\: 1]$
This problem here

The maximum value is $1/3$. First, the integral is equal to $1/3$ when $y=1$, so the maximum is at least that large. To see that the integral is never bigger than $1/3$, note that $$\sqrt{x^4+(y-y^2)^2}\leq x^2 + y-y^2\quad \text{for $y\in [0,1]$},$$
and therefore that
\begin{align*}
\int_0^y \sqrt{x^4+(y-y^2)^2}\,dx &\leq \int_0^y(x^2+y-y^2)\,dx = y^2-{2\over3}y^3.\tag{1}
\end{align*}
But $g(y) = y^2-{2\over3}y^3$ is increasing on $[0,1]$ since $g'(y) = 2(y-y^2)\geq0$ there. So $g(y)\leq g(1) = 1/3$ for all $y\in[0,1]$, and the proof is finished by $(1)$.
