Let $$f(x)= \int_{0}^{x} \sqrt{(x^2-x)^2+y^4} \,dy, $$ and $x \in [0,1].$ What is the maximum value of $f(x)?$ At the endpoints $f$ is equal to $0$ and $1/3$ respectively. I tried to find the stationary points from $f'(x)=0$ using the Leibniz integral rule and concluded that on the interval $[0,\frac{1}{2}]$ the only stationary point is $0$. But I got stuck with the case $x \in [\frac{1}{2}, 1]$. Could someone help me with the latter case or provide some better methods? Thanks in advance.
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I take a break from grading my calculus exams and make a silly mistake like that. How embarrasing. Thanks for the correction @MaximilianJanisch – James Dec 21 '21 at 22:51
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@James Haha it‘s all good, happy correcting and merry Christmas – Maximilian Janisch Dec 21 '21 at 22:52
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A numerical test seems to show that $f$ increases monotonically. The maximum would be $1/3$ at $x=1$. Actually, even $f(x)/x$ seems to be increasing. – Gribouillis Dec 21 '21 at 23:06
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Also: here, here, and here – bjorn93 Dec 22 '21 at 10:18