Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$ Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx  $ for $y\in[0,1].$
I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but that also is not working.Can someone please help me in solving this question?

Let
$$
I(y) = \int_0^y \sqrt{x^4 + (y-y^2)^2}dx
$$
Let's find $y$ such that $dI/dy=0$. By Leibniz integral rule we have
$$
\frac{dI}{dy} = \int_0^y \frac{\partial}{\partial y}\sqrt{x^4 + (y-y^2)^2}dx + \sqrt{y^4 + (y-y^2)^2}
$$
But $dI/dy>0$ for $y>0$ ($dI/dy=0$ at $y=0$). So, maximum is reached for $y=1$:
$$
I(1)=\int_0^1 x^2\,dx = 1/3
$$
Here is plot of $72I(y)$:

EDIT
(Answer to @PhoemueX's comment)
We should to prove that $dI/dy>0$. We have
$$
\frac{dI}{dy} = \int_0^y \frac{(1-2 y) (y-y^2)}{\sqrt{x^4+\left(y-y^2\right)^2}}dx + \sqrt{y^4 + (y-y^2)^2}.
$$
It's trivial for $y\le 1/2$ (both terms are positive). For $1/2 < y \le 1$
$$
\frac{dI}{dy} > -\int_0^y \frac{(2 y-1) (y-y^2)}{\sqrt{(y-y^2)^2}}dx + \sqrt{y^4 + (y-y^2)^2} = \\= \sqrt{y^4 + (y-y^2)^2}-(2y-1)y\sqrt{y-y^2}=\\=
\sqrt{y^4 + (y-y^2)^2}-\sqrt{y^2(2y-1)^2(y-y^2)}
$$
But
$$
y^4 + (y-y^2)^2>y^2(2y-1)^2(y-y^2)
$$
