Elastic potential energy derivation

For a string If $F_1=0$ then $x_1=0$ , $F_2=F$ then $x_2=x$ Elastic potential energy = work done = $\frac{F_1+F_2} {2} \cdot x = (0+F)2 \cdot x= \frac{Fx} {2} = -\frac{1} {2} Kx^2 $ In this derivation why should we take average force $\frac{F_1+F_2} {2} $ and why we are not taking average displacement of wire $\frac{x_1 +x_2} {2} $

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    $\begingroup$ Welcome to Physics SE! You're highly encouraged to use LaTeX in your questions. $\endgroup$ – Mauro Giliberti Jun 13 '19 at 5:58

You can take the average position as well, the result don't change. Be aware that in those formulas what's important isn't really the displacement $x$, but the change in displacement $\Delta x$. In this case they are the same only because $x_1=0$.

$\frac{x_1+x_2}{2} \cdot F =\frac{0+x}{2} \cdot F =\frac{Fx}{2} $


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