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?
 A: 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
$$
y^4 + (y-y^2)^2>y^2(2y-1)^2(y-y^2)
$$
A: As observed by @MichaelGaluza, by Leibniz's rule, it suffices to show
$$
\sqrt{y^{4}+\left(y-y^{2}\right)^{2}}+\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq0.
$$
Note that $y-y^{2}=y\left(1-y\right)>0$ for $y\in\left(0,1\right)$.
Now, for $y\leq\frac{1}{2}$, we have $\left(1-2y\right)\left(y-y^{2}\right)\geq0$,
so that the claim is trivial.
Thus, we can assume $y\in\left(\frac{1}{2},1\right)$. In this case,
we have $\left(1-2y\right)\left(y-y^{2}\right)<0$. Furthermore,
$$
\int_{0}^{y}\frac{1}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\leq\int_{0}^{y}\frac{1}{\sqrt{\left(y-y^{2}\right)^{2}}}\,{\rm d}y=\frac{y}{y-y^{2}}=\frac{1}{1-y}.
$$
Multiplying by $\left(1-2y\right)\left(y-y^{2}\right)<0$ yields
$$
\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq\frac{\left(1-2y\right)\left(y-y^{2}\right)}{1-y}=\left(1-2y\right)y,
$$
so that it suffices to show
\begin{align*}
 & \sqrt{y^{4}+\left(y-y^{2}\right)^{2}}\geq\left(2y-1\right)y\\
\left(\text{since }\left(2y-1\right)y\geq0\text{ because }y>\frac{1}{2}\right)\Longleftrightarrow & y^{4}+\left(y-y^{2}\right)^{2}\geq\left(2y-1\right)^{2}y^{2}\\
\Longleftrightarrow & y^{4}+y^{2}-2y^{3}+y^{4}=y^{4}+\left(y-y^{2}\right)^{2}\geq\left(4y^{2}-4y+1\right)y^{2}=4y^{4}-4y^{3}+y^{2}\\
\Longleftrightarrow & 0\geq2y^{4}-2y^{3}=2y^{3}\left(y-1\right)\\
\left(\text{since }y>0\right)\Longleftrightarrow & y-1\leq0\\
\Longleftrightarrow & y\leq1,
\end{align*}
which is true. This proves that the derivative is indeed nonnegative,
so that the maximum is attained at $y=1$.
