Power as a Function of Distance I am interested in trying to understand the meaning of the integral of power with respect to distance.
In the case of force, we have the two formulae: $\int F \, \operatorname{d}\!t = \Delta p$ and $\int F \, \operatorname{d}\!x = \Delta E_k$.
I understand that power is closely linked to time and work by the relationship $P = \frac{\operatorname{d}\!W}{\operatorname{d}\!t}$.
However, imagine a body moving continuously from the left to the right. In such a case, we can express displacement as a one-to-one function of time, and time as a one-to-one function of displacement. We could place regular markers along the road side and take power readings as the body passes each of these markers. Assuming the distance between markers is $\Delta s$ and then letting $\Delta s \to 0$  gives power as a function of displacement. We may then ask: What does the integral of this function mean?
$$\int_{s_1}^{s_2} P \, \operatorname{d}\!x = \operatorname{F}(s_1,s_2)$$
Addition:
I've tried to play with the calculus a bit. I'm not sure if this helps:
$$P = \frac{\operatorname{d}\!W}{\operatorname{d}\!t} = \frac{\operatorname{d}\!x}{\operatorname{d}\!t} \frac{\operatorname{d}\!W}{\operatorname{d}\!x} = v\frac{\operatorname{d}\!W}{\operatorname{d}\!x}\implies \int P \, \operatorname{d}\!x = \int v\frac{\operatorname{d}\!W}{\operatorname{d}\!x} \, \operatorname{d}\!x = \int v \, \operatorname{d}\!W$$
Has anyone seen a velocity-work graph? If so, what does the area underneath it represent?
 A: Maybe this can help. Consider a body accelerating under constant power from the initial conditions $t=0$, $x=x_1$ and $v=v_1$.
$$ a(v) = \frac{P}{m v} $$
$$ t(v) = \int \frac{1}{a(v)}\,{\rm d} v  = \int \frac{m v}{P} \,{\rm d}v = \frac{m}{P} \left( \frac{v^2}{2} - \frac{v_1^2}{2} \right) $$
$$ v(t) = \sqrt{ \frac{2 P t}{m} + v_1^2}  $$
$$ x(v) = \int \frac{v}{a(v)}\,{\rm d} v = \int \frac{m v^2}{P} \,{\rm d}v =x_1 + \frac{m}{P} \left( \frac{v^3}{3} - \frac{v_1^3}{3} \right)$$
$$ v(x) = \sqrt[3]{\frac{3 P x}{m} + v_1^3} $$
$$ x(t) = x(v(t)) =x_1 + \frac{m}{P} \left(\frac{1}{3} \left( \frac{2 P t}{m} + v_1^2 \right)^{\frac{3}{2}} - \frac{v_1^3}{3} \right) $$
So it seems the quantity you are looking for is 
$$ \int P \,{\rm d}x= P \int {\rm d}x=P\, x(v) $$
for constant power.
In general, power is a function of speed, so you can say
$$ \int P(v) \,{\rm d}x = \int P(v) \frac{v}{a} \,{\rm d}v = \int m v^2 \,{\rm d}v = \frac{1}{3} m \left( v^3 - v_1^2 \right) $$
which appears to be scaling factor for $x(v)$. In fact, you can state that the average power within a distance is
$$ P_{ave} = \frac{ \int P \,{\rm d} x}{\int \,{\rm d} x} = \frac{ \frac{1}{3} m \left( v^3 - v_1^3 \right)}{x-x_1} $$
So you can take a measurement of the distance traveled ($x_1 \rightarrow x$) and the speed change ($v_1 \rightarrow v$) and estimate the average power needed to reach those numbers. Same way that people use the trap speed of a 1/4 mile drag race to estimate power.
A: What you did so far is correct.
Use the fact that $ W = 1/2 \cdot m \cdot v^{2}$.
Here are the final steps from your work to the solution, use integration by parts:
$$ \int v \cdot dW = 1/2 \cdot m \cdot v^{3}|_{v_1}^{v_2} - \int_{v_1}^{v_2} 1/2 \cdot m \cdot v^{2} =1/3 \cdot m \cdot [v_2^3 - v_1^3] $$
