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?

Re: Finding a function through its derivatives integration over a finite distance

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**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 $\displaystyle E_0- 0= E_0$.

Re: Finding a function through its derivatives integration over a finite distance

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**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.