TITLE: Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$?
QUESTION [5 upvotes]: How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?

REPLY [4 votes]: Since $g(y)=y(1-y)$ is increasing over $I=\left[0,\frac{1}{2}\right]$, so it is $f$. Assume now $y\in\left[\frac{1}{2},1\right]$. 
We have:
$$ f'(y)=-\int_{0}^{y}\frac{(y-y^2)(2y-1)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2},\tag{1}$$
but in virtue of the mean value theorem:
$$ \int_{0}^{y}\frac{(y-y^2)(2y-1)}{\sqrt{x^4+(y-y^2)^2}}dx \leq y(2y-1)\tag{2}$$
so it is sufficient to prove that:
$$ y^2+(1-y)^2 \geq (2y-1)^2 \tag{3}$$
for any $y\in\left[\frac{1}{2},1\right]$, to have that $f$ is increasing over $[0,1]$. However, $(3)$ is equivalent to $y(1-y)\geq 0$, that is trivial. This gives that the maximum of $f(y)$ over $[0,1]$ is attained in $y=1$, and the maximum is $\frac{1}{3}$.