Find the maximum of $f(x) =\int_0^x\sqrt{(x^2-x)^2+y^4}\, dy, x\in[0,1].$ Find the maximum of $$f(x) =\int_0^x\sqrt{(x^2-x)^2+y^4}\, dy, x\in[0,1].$$


I am generally familiar with the process of how to find the maximum of a function on a given interval, however I really don't even know where to start with this one, any hints would be greatly appreciated!
Mathematica 12.0 answers

Maximize[{Integrate[Sqrt[(x^2 - x)^2 + y^4], {y, 0, x}],  x &gt;= 0 &amp;&amp; x &lt;= 1}, x]


$$\left\{\frac{\Gamma \left(\frac{3}{4}\right)}{4 \Gamma \left(\frac{7}{4}\right)},\{x\to 1\}\right\} $$
First, we have to get the derivarive of $ f $. 

So, let’s use the Fundamental Theorem of Calculus in order to find it:

$$f’(x) = \sqrt{(x^2 - x)^2 + x^4} \cdot 1
= |x| \sqrt{2x^2-2x+1}$$ 

Since the interval is positive, it’s just $x $ times that square root.

Therefore, we must avaluate at $f’=0$ or where the function “does not exist”. However, the function is well behaved at the whole real line, since the roots are complex conjugates.
$$f’(x) = x \sqrt{2x^2 -2x +1} = 0.$$
Therefore, $ x = 0, f(0)=0$.

Now, do not forget the boundaries of the interval: 0, 1. Since, we have already gotten the result above:

$f(1) =$ something that I will finish further, I’m kinda busy right now. But that’s the idea.