- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.

- Jan 29, 2012

- 1,151

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 [tex]f(y)= \int_a^y g(x,y) dx[/tex] then [tex]df/dy= g(y,y)[/tex]. So you should find y such that [tex]\sqrt{y^4+ (y- y^2)^2}= 0[/tex]. (If that y is not between 0 and 1 then look at the value at 0 and 1.)Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.

- Admin
- #3

- Mar 5, 2012

- 9,006

That doesn't look quite right.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 [tex]f(y)= \int_a^y g(x,y) dx[/tex] then [tex]df/dy= g(y,y)[/tex]. So you should find y such that [tex]\sqrt{y^4+ (y- y^2)^2}= 0[/tex]. (If that y is not between 0 and 1 then look at the value at 0 and 1.)

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:

$$g(y,y) = y+y = 2y \ne 3y$$

- Jan 17, 2013

- 1,667

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

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

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

So the FTC doesn't apply directly here.

- Jan 31, 2012

- 253

- Jan 17, 2013

- 1,667

That integral doesn't seem to be solvable in terms of elementary functions. The W|A returns an elliptic integral.

- Thread starter
- Admin
- #7

- Feb 14, 2012

- 3,812

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

Hmm...that's not quite right,That integral doesn't seem to be solvable in terms of elementary functions.

Having said so, I will only post the solutions days later with the hope that others may want to take a stab at it.

- Jan 17, 2013

- 1,667

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.Hmm...that's not quite right,Zaid!

- Thread starter
- Admin
- #9

- Feb 14, 2012

- 3,812

Most probably that I am wrong,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.

- Moderator
- #10

- Feb 7, 2012

- 2,740

Find the exact maximum value of $\int_0^y \sqrt{x^4+(y-y^2)^2} dx$ for $0 \le y \le 1$.

- Thread starter
- Admin
- #11

- Feb 14, 2012

- 3,812

Well done,