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How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?

Aaron Maroja
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piteer
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4 Answers4

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Since $g(y)=y(1-y)$ is increasing over $I=\left[0,\frac{1}{2}\right]$, so it is $f$. Assume now $y\in\left[\frac{1}{2},1\right]$.

We have: $$ f'(y)=-\int_{0}^{y}\frac{(y-y^2)(2y-1)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2},\tag{1}$$ but in virtue of the mean value theorem: $$ \int_{0}^{y}\frac{(y-y^2)(2y-1)}{\sqrt{x^4+(y-y^2)^2}}dx \leq y(2y-1)\tag{2}$$ so it is sufficient to prove that: $$ y^2+(1-y)^2 \geq (2y-1)^2 \tag{3}$$ for any $y\in\left[\frac{1}{2},1\right]$, to have that $f$ is increasing over $[0,1]$. However, $(3)$ is equivalent to $y(1-y)\geq 0$, that is trivial. This gives that the maximum of $f(y)$ over $[0,1]$ is attained in $y=1$, and the maximum is $\frac{1}{3}$.

Jack D'Aurizio
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By differentiation under the integral sign we get \begin{align*}\left(\int_0^y\sqrt{x^4+(y-y^2)^2}\,dx\right)'&=\int_0^y\frac{2(y-y^2)(1-2y)}{2\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}\\&=\sqrt{y^4+(y-y^2)^2}+\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx,\end{align*} which is positive for $y\in(0,\frac12)$. Else suppose $y\in(\frac12,1)$, then it's positive if $$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx=-\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx<\sqrt{y^4+(y-y^2)^2}=y\sqrt{y^2+(1-y)^2}.$$ Clearly, $$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx<y\cdot\frac{(2y-1)(y-y^2)}{\sqrt{(y-y^2)^2}}=y(2y-1),$$ so it remains to check that $$2y-1\le\sqrt{y^2+(1-y)^2}\iff 0\le y^2+(1-y)^2-(2y-1)^2=-2y^2+2y=2y(1-y)$$ So we can conclude the maximum is attained for $y=1$, where it's $$\int_0^1\sqrt{x^4}\,dx=\int_0^1 x^2\, dx=\frac13.$$

user2345215
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  • Whoops, I have just posted a solution that is almost identical to yours, I did not notice you wrote it. Hope you don't mind. (+1), anyway. – Jack D'Aurizio Dec 27 '14 at 20:46
  • @JackD'Aurizio Well, that sometimes happens. As long as it's independent, it's perfectly fine, so have a +1 too :) – user2345215 Dec 27 '14 at 20:54
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HINT:

By taking derivative with respect to $y$ and setting it to be equal to $0$ we have : $$\int_{0}^{y}\frac{(y-y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}=0$$ Hence: $$\int_{0}^{y}\frac{dx}{\sqrt{x^4+(y-y^2)^2}}dx=\frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)}$$

CLAUDE
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$$\begin{align}\frac{d}{dy}(f(y)) &= \frac{d}{dy} \Bigg(\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx\Bigg) \\&= \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} . 1\end{align}$$

Now $$\begin{align}&\int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} = 0 \\&\Rightarrow \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx = - \sqrt{y^4+(y-y^2)^2} \\&\Rightarrow \int_{0}^{y}\frac{1}{\sqrt{x^4+(y-y^2)^2}} dx = \frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)} \end{align}$$

Check the sign of $\frac{d}{dy}f(y)$ to see if the function $f(y)$ is decreasing, increasing.

Aaron Maroja
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