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 $y\in \left[0,1\right]$?
 A: 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 $\frac{1}{3}$.
A: 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
$$\int_0^1\sqrt{x^4}\,dx=\int_0^1 x^2\, dx=\frac13.$$
A: HINT:
By taking derivative with respect to $y$ and setting it to be equal to $0$ we have :
$$\int_{0}^{y}\frac{(y-y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}=0$$
Hence:
$$\int_{0}^{y}\frac{dx}{\sqrt{x^4+(y-y^2)^2}}dx=\frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)}$$
A: $$\begin{align}\frac{d}{dy}(f(y)) &= \frac{d}{dy} \Bigg(\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx\Bigg) \\&= \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} . 1\end{align}$$
Now $$\begin{align}&\int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} = 0 \\&\Rightarrow \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx = - \sqrt{y^4+(y-y^2)^2} \\&\Rightarrow \int_{0}^{y}\frac{1}{\sqrt{x^4+(y-y^2)^2}} dx = \frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)} \end{align}$$
Check the sign of $\frac{d}{dy}f(y)$ to see if the function $f(y)$ is decreasing, increasing.
