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
 A: 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 $(1)$.
A: I'm sorry but I think that what DanielV said is false, mainly because we don't have $\int^y f(x)\text{d}x$ but $\int^y f(x,y)\text{d}x$
For example, if we take the function $f(x,y) = 1$ if $0 \leqslant y \leqslant \frac 12$ and $f(x,y) = 0$ if $\frac 12 < y \leqslant 1$. Then clearly $f(x,y) \geqslant 0$ but $$F\left(\frac 12\right) =\frac 12 > 0 = F(1)$$
(if we define $$F(y) = \int_0^y f(x,y)\text{d}x$$)
I haven't really thought about the exerice though... The main issue is that we have an occurrence of $y$ in the function to integrate and in the interval on which we integrate. Maybe try to get rid of one ?
A: This is not an answer for an homework problem but rather a curiosity you could be interested by.   
The antiderivative has an analytical expression which involves a rather complex elliptic integral of the first kind. However, the integral simplifies and, for 0 < y <1, write  
y^2 (1 - y) Hypergeometric2F1[-1/2, 1/4, 5/4, -y^2/(1 - y)^2]  
Over the range 0 < y < 1, this function is continuously increasing (there are two inflexion points very close to y=0.4 and y=0.7). For y=1, the limit value of the function is 1/3.
