MHB Maximize Integral: Find Exact Value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$

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The discussion focuses on finding the maximum value of the integral $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$. Participants emphasize the need to differentiate the integral to locate critical points, noting that the derivative involves evaluating the integrand at the upper limit. It is concluded that the integral does not yield an elementary anti-derivative and may instead result in an elliptic integral. Using inequalities, it is shown that the maximum value occurs at $y=1$, yielding a maximum of $1/3$. The final consensus is that the maximum value of the integral on the specified interval is $1/3$.
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Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
 
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anemone said:
Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if f(y)= \int_a^y g(x,y) dx then df/dy= g(y,y). So you should find y such that \sqrt{y^4+ (y- y^2)^2}= 0. (If that y is not between 0 and 1 then look at the value at 0 and 1.)
 
HallsofIvy said:
I presume you know that to determine maximum or minimum values for a differentiable function, you set the derivative equal to 0. You should also know that if f(y)= \int_a^y g(x,y) dx then df/dy= g(y,y). So you should find y such that \sqrt{y^4+ (y- y^2)^2}= 0. (If that y is not between 0 and 1 then look at the value at 0 and 1.)

That doesn't look quite right.

Suppose $g(x,y) = x+y$.

Then:
\begin{aligned}\frac{d}{dy}\int_0^y g(x,y)dx
&= \frac{d}{dy}\int_0^y (x+y)dx \\
&= \frac{d}{dy}\left(\frac 1 2 x^2 + xy \Big|_0^y\right) \\
&= \frac{d}{dy}\left(\frac 3 2 y^2\right) \\
&= 3y
\end{aligned}
But:
$$g(y,y) = y+y = 2y \ne 3y$$
 
We cannot differentiate with respect to $y$ while it is inside the integral so we have first to separate them then

$$\int^y_0 (x+y)dx = \int^y_0 xdx +y\int^y_0 dx $$

Now if we differentiate and according to the product rule we have

$$\frac{df}{dy}\int^y_0 (x+y)dx=\frac{df}{dy} \left( \int^y_0 xdx +y\int^y_0 dx \right) =y +\int^y_0 dx+y = 3y $$

So the FTC doesn't apply directly here.
 
If $ \displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $ \displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.
 
Random Variable said:
If $ \displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $ \displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.

That integral doesn't seem to be solvable in terms of elementary functions. The W|A returns an elliptic integral.
 
Thank you all for the feedback!

Random Variable said:
If $ \displaystyle f(y)= \int_a^y g(x,y) \ dx$, then $ \displaystyle \frac{df}{dy}= \int_{a}^{y} g_{y}(x,y) \ dx + g(y,y)$.

I believe you're right, and if one wants to attack the problem using this definition, then that would be welcome!

ZaidAlyafey said:
That integral doesn't seem to be solvable in terms of elementary functions.

Hmm...that's not quite right, Zaid! But speaking of more advanced integration field, you and the rest of the members are the experts, not me. I want to let you know that one of the solutions that I have solved it through the inequality route and I actually can't wait to share it with MHB! :)

Having said so, I will only post the solutions days later with the hope that others may want to take a stab at it.
 
anemone said:
Hmm...that's not quite right, Zaid!

Well, I don't know whether I am messing something but I still cannot think how to find an elementary anti-derivative for that integral.
 
ZaidAlyafey said:
Well, I don't know whether I am messing something but I still cannot think how to find an elementary anti-derivative for that integral.

Most probably that I am wrong, Zaid. Please don't take what I said seriously because really I am no comparison to you when it comes to the territory of advance integration.
 
  • #10
anemone said:
Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.
[sp]Let $$f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.$$ My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be $$f(1) = \int_0^1 x^2\, dx = 1/3.$$ Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that $$f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.$$ But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$[/sp]
 
  • #11
Opalg said:
[sp]Let $$f(y) = \int_0^y \sqrt{x^4+(y-y^2)^2} dx.$$ My instinct is that the maximum value of $f$ over the interval $[0,1]$ must be $$f(1) = \int_0^1 x^2\, dx = 1/3.$$ Following anemone's hint about using an inequality, it occurred to me that if $a,b \geqslant 0$ then $\sqrt{a^2+b^2} \leqslant a+b.$ Applying that with $a=x^2$ and $b= y-y^2$, you see that $$f(y) \leqslant \int_0^y(x^2 + y - y^2)\,dx = \tfrac13y^3 + y^2 - y^3 = y^2 - \tfrac23y^3.$$ But $y^2 - \tfrac23y^3$ is an increasing function on the interval $[0,1]$, with a maximum value $1/3$ when $y=1$. Therefore the maximum value of $f(y)$ is $f(1) = 1/3.$[/sp]

Well done, Opalg, and thanks for participating! (Sun)
 
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