What would the "correct" way of doing this integral be?

[EddiePhys]

v(x(t)), where v represents velocity and is a function of position which is a function of time.

I have the equation: v dv/dx = 20x + 5 and want to solve for velocity. The way our professor solved it was by multiplying both sides by dx and integrating => ∫v dv = ∫20x+5 dx. I know doing this is non-rigorous since dv/dx can't be treated as a fraction. So how would I "correctly" do this?

Generally ∫ ƒ(y) dy/dx dx = ∫ f(u) du by substitution or the "reverse" chain rule since dy/dx is the derivative of the composite function y inside f In my case however, ∫ v dv/dx , dv/dx is not the derivative of the composite inside v. How would I solve for v without doing something like multiplying both sides by dx?

 

[Orodruin]

You are just intrgrating over t. Note that v dt = (dx/dt) dt = dx so you can just apply substitution of variables to x instead of t.

 

[EddiePhys]

You are just intrgrating over t. Note that v dt = (dx/dt) dt = dx so you can just apply substitution of variables to x instead of t.

I made a few mistakes initially sorry. Corrected them. Please look at the post again.

 

[Orodruin]

This does not really change the principle, it is still just integrating both sides with respect to a variable, in this case x, and changing the integral parametrisation.

 

[EddiePhys]

This does not really change the principle, it is still just integrating both sides with respect to a variable, in this case x, and changing the integral parametrisation.

v dv/dx = 20x + 5 If I want to integrate both sides with respect to x, how do I do it?
The RHS would be pretty straightforward.

But how do I integrate this:
∫ v(x) dv/dx dx without "cancelling" the dx's

 

[Orodruin]

You are not "cancelling" the dx. You are using the normal form for a change of variables.

 

[EddiePhys]

You are not "cancelling" the dx. You are using the normal form for a change of variables.

But for change of variables, the "inner" or composite function's derivative has to be present outside like: ∫f(x(t)) dx/dt dx = ∫f(y) dy.
In this case ∫ v(x) dv/dx dx, dv/dx is not the derivative of the composite function but it is the derivative of v(x)

 

[Orodruin]

You are changing variables to v ... Also, your expression for change of variables is wrong. Fix it with replacing dx by dt and it is exactly what you have.

 

[Orodruin]

Of course, you could also just note that v dv/dx = d(v^2)/dx / 2 and integrate a total derivative.

 

[EddiePhys]

You are changing variables to v ... Also, your expression for change of variables is wrong. Fix it with replacing dx by dt and it is exactly what you have.

In the change of variable, ∫ g(f(x)) df/dx dx = ∫ g(u) du. But my case is more analogous to ∫ g(f(x)) dg/df df since I have ∫ v(x(t)) dv/dx dx

 

[Orodruin]

In the change of variable, ∫ g(f(x)) df/dx dx = ∫ g(u) du. But my case is more analogous to ∫ g(f(x)) dg/df df since I have ∫ v(x(t)) dv/dx dx

No it is not. You just have g(f) = f.
The fact that v and x generally depend on t here is irrelevant.

 

[EddiePhys]

No it is not. You just have g(f) = f.
The fact that v and x generally depend on t here is irrelevant.

Alright got it thanks