Energy stored in spring stretched diagonally

Hi, just want some clarification on this please. We have a spring on natural length l which is held vertically. The end is then moved horizontally so that the spring is stretched diagonally. The question is to find the stored energy when the end is at position x.

I said that the length of the spring, L=(l^2+x^2)^0.5, which for small displacements can be approximated by l(1+x^2/2l^2). The extension is then x^2/2l. The force is thus kx^2/2l. Here is what I don't get. The solutions say that the stored energy is
$[br]\int \dfrac{kx^2}{2\ell}\, \mathrm{d}x[br]$.
Why is the integration with respect to x? Shouldn't a tiny increase in energy, dW=F*dL, where dL is a tiny extension of the spring? I can't understand why the solutions write dW=kdx

in dW=F*dL the force is kx^2/2l, and it should be dx not dL,because L is the length not the extension,the extension is x, if you drew a graph of force and extension,the area under the graph is the work done, so dW=Fdx

L is a constant, x is the variable.
The extension you say is x²/2L
The force is kx²/2L
The stored energy dW from moving the spring dx is the integral Fdx

But x is the distance the end of the spring is moved horizontally, the spring extends in the diagonal direction

This doesn't make sense to me though, dx isn't the extension of the spring so why are we integrating with respect to it?

L is a constant, x is the variable.
The extension, e, you say is x²/2L
The force is ke
The force is kx²/2L

The stored energy dW from moving the spring dx is the integral Fdx

Because you want the energy stored in terms of x

Hi, just want some clarification on this please. We have a spring on natural length l which is held vertically. The end is then moved horizontally so that the spring is stretched diagonally. The question is to find the stored energy when the end is at position x.

I said that the length of the spring, L=(l^2+x^2)^0.5, which for small displacements can be approximated by l(1+x^2/2l^2). The extension is then x^2/2l. The force is thus kx^2/2l. Here is what I don't get. The solutions say that the stored energy is
$[br]\int \dfrac{kx^2}{2\ell}\, \mathrm{d}x[br]$.
Why is the integration with respect to x? Shouldn't a tiny increase in energy, dW=F*dL, where dL is a tiny extension of the spring? I can't understand why the solutions write dW=kdx

we should multiply the integral by (dx/de)x(de/dx)

So just x then?

no, in your integral it should be W=S(kx^2/2L)de where e is the extension of the string,not x,
then multipliy it by (dx/de)(de/dx) so
W=S(kx^2/2L)de times (dx/de)(de/dx), the de's cancel with de(dx/de)and you get
W=S(kx^2/2L)dx times (de/dx) but you know the extension e=x^2/2L
so you can find de/dx

so just multiply W=S(kx^2/2L)dx by de/dx and you will get the work done in terms of dx which i got was W=S(kx^3/2L^2)dx

EDIT: it was meant to put so you can find de/dx, not dx/de just changed it now

That gives the incorrect answer according to the solutions

shouldnt it be Fde?

You guys are making it more complicated than it needs to be,

The problem is about how much energy is stored in the spring when the end is moved horizontally a distance x.
So you want at the end the answer in terms of x.
That is, if I pull the spring end horizontally a distance x, I plug x into the formula to get the energy stored.

The energy stored depends on the extension of the spring δe, so you need first mathematically to find the extension in terms of x.
You then have δe in terms of δx

When you have that, mathematically, you integrate the formula with respect to x, as x is now the variable, not the extension.

The resulting formula gives you the energy stored as a function of x. Which is what you want.
There is no point in integrating wrt e as you don't know e. You only know the value of x, which e depends on.

You guys are making it more complicated than it needs to be,

The problem is about how much energy is stored in the spring when the end is moved horizontally a distance x.
So you want at the end the answer in terms of x.
That is, if I pull the spring end horizontally a distance x, I plug x into the formula to get the energy stored.

The energy stored depends on the extension of the spring δe, so you need first mathematically to find the extension in terms of x.
You then have δe in terms of δx

When you have that, mathematically, you integrate the formula with respect to x, as x is now the variable, not the extension.

The resulting formula gives you the energy stored as a function of x. Which is what you want.
There is no point in integrating wrt e as you don't know e. You only know the value of x, which e depends on.