# Finding a function through its derivatives integration over a finite distance

• Jun 17th 2013, 02:28 PM
fysikbengt
Finding a function through its derivatives integration over a finite distance
I dont know if this Counts as a differential equation, but here it is. $\displaystyle \int_0^d \frac{\partial E}{\partial x}dx=E_0$. The condition is that $\displaystyle \frac{\partial E}{\partial x}>0$ in the interval [0,d] and that $\displaystyle E(0)=0$ and $\displaystyle E(d)=E_0$. I can see that a solution is $\displaystyle E=xE_0 /d$ but I dont know how to find this without guessing. Is there a simple method to employ?
• Jun 17th 2013, 04:54 PM
HallsofIvy
Re: Finding a function through its derivatives integration over a finite distance
Quote:

Originally Posted by fysikbengt
I can see that a solution is $\displaystyle E=xE_0 /d$.

I'm sorry to hear that because it is completely wrong. You seem to be assuming that E is the constant, $\displaystyle E_0$, which is certainly not true. I'm not sure why you write partial derivatives when you only have the one variable, x. One part of the "Fundamental Theorem of Calculus" say that $\displaystyle \int_0^d\frac{dE}{dx}dx= E(d)- E(0)$. Since we are told that $\displaystyle E(0)= E_0$ and E(0)= 0, that is $\displaystyle E_0- 0= E_0$.
• Jun 18th 2013, 09:04 AM
fysikbengt
Re: Finding a function through its derivatives integration over a finite distance
Quote:

Originally Posted by HallsofIvy
I'm sorry to hear that because it is completely wrong. You seem to be assuming that E is the constant, $\displaystyle E_0$, which is certainly not true. I'm not sure why you write partial derivatives when you only have the one variable, x. One part of the "Fundamental Theorem of Calculus" say that $\displaystyle \int_0^d\frac{dE}{dx}dx= E(d)- E(0)$. Since we are told that $\displaystyle E(0)= E_0$ and E(0)= 0, that is $\displaystyle E_0- 0= E_0$.

Yes, I used an equality by definition so it had to go in circles. But you were good help anyway since I now had to make a more general approach to my original problem doing a variable substitution from x to E to solve it. It was a much shorter and neater proof.