maximum value of \(f(y)\)

[jacks]

Maximum value of $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx\;,$$ Where \(y\in \left[0,1\right]\)

[Tolaso J Kos]

The integrand is strictly positive for \( 0<x \leq 1 \) and \( 0 \leq y \leq 1 \). So \( f(y) >0 \) for \( 0<z\leq 1 \).

It should be pointed out that we cannot get the integral into a closed form except of \( y=0, \; y=1 \). But we have a strong huntch that the maximum must be the higher of the two values we can actually calculate. It is also easy to see that \( \displaystyle f(1)=\frac{1}{3} \).

It is easy to see that \( f \) is differentiable. Then: $$f'(y)=\sqrt{y^4+\left [y^2\left ( 1-y^2 \right ) \right ]}+y\left ( 1-y \right )\left ( 1-2y \right )\int_{0}^{y}\left ( x^4+y^2\left ( 1-y^2 \right ) \right )^{-1/2}\,dx$$
The integrand is positive so \( f'(y)>0 \;\; \forall y \in (0, 1/2] \). The integrand is less than \( \displaystyle \frac{1}{\sqrt{y^2\left ( 1-y^2 \right )}}=\frac{1}{y\left ( 1-y \right )} \) , so the integral is less than \( \displaystyle \frac{1}{1-y} \). For \( y>1/2 \) we have: $$f'(y)>\sqrt{y^4+y^2\left ( 1-y^2 \right )}-y\left ( 2y-1 \right )$$
Squaring we see that: \( \displaystyle y\left ( 2y-1 \right )<\sqrt{y^4+y^2\left ( 1-y^2 \right )} \) for \( 1/2<y<1 \) and with that the proof comes to an end!