How can I solve something like this?

[Uggohe]

Hi, I was wondering how can I solve an integral with a differential function
[MATH]\int x^2 dx^2 [/MATH]

 

[Dr.Peterson]

Hi, I was wondering how can I solve an integral with a differential function
[MATH]\int x^2 dx^2 [/MATH]

Please provide some context. Where did this integral come from? What does it mean? Also, what have you tried?

 

[Uggohe]

Hi, thanks for your reply!
I just invented that function at the moment, I was trying to understand how can I solve an integral where the differential is a function and not just a dx.
I was studying some physics stuff and I came across [MATH]v\cdot dv[/MATH], thus I started wondering what really means to multiply a function for the differential of the function itself.
What does [MATH]f(x)\cdot df(x)[/MATH]means?
Do I have to integrate it to find the result?

 

[Dr.Peterson]

Hi, thanks for your reply!
I just invented that function at the moment, I was trying to understand how can I solve an integral where the differential is a function and not just a dx.
I was studying some physics stuff and I came across [MATH]v\cdot dv[/MATH], thus I started wondering what really means to multiply a function for the differential of the function itself.
What does [MATH]f(x)\cdot df(x)[/MATH]means?
Do I have to integrate it to find the result?

That helps a lot; you didn't mean what you wrote (or rather, what I read)! I was thinking of "\(dx^2\)" as "\((dx)^2\)", or "\(dxdx\)" as in \(\frac{d^2y}{dx^2}\). That would have made this a double integral, and your notation would be quite odd.

You are thinking instead of \(d(x^2)\), the differential of the function \(y = x^2\). In this sense, \(dy = 2x dx\). So \(\int x^2d(x^2)=\int x^2\cdot 2x dx=\int 2x^3 dx=\frac{1}{2}x^4 + C\).

And in fact, it could have been expressed simply as \(\int y dy=\frac{1}{2}y^2 + C=\frac{1}{2}x^4 + C\).

You may well get an answer from a physics perspective; but it may help if you tell us more about the context in which you found this.

 

[Uggohe]

That helps a lot; you didn't mean what you wrote (or rather, what I read)! I was thinking of "\(dx^2\)" as "\((dx)^2\)", or "\(dxdx\)" as in \(\frac{d^2y}{dx^2}\). That would have made this a double integral, and your notation would be quite odd.

You are thinking instead of \(d(x^2)\), the differential of the function \(y = x^2\). In this sense, \(dy = 2x dx\). So \(\int x^2d(x^2)=\int x^2\cdot 2x dx=\int 2x^3 dx=\frac{1}{2}x^4 + C\).

And in fact, it could have been expressed simply as \(\int y dy=\frac{1}{2}y^2 + C=\frac{1}{2}x^4 + C\).

You may well get an answer from a physics perspective; but it may help if you tell us more about the context in which you found this.

Well, thank you, that cleared my ideas a lot.
But I am still not getting the sense of writing something like [MATH]g(x) = f(x)dx[/MATH].
Is [MATH]g(x)[/MATH] equal to [MATH]\int f(x)dx[/MATH]?
Or does it is have a different meaning?
-
Ok, I've been thinking about it pretty much and maybe it makes sense now.
That just means what I see, a function for the differential of a variable, which actually isn't an integral.
The integration operation "automatically" multiplies the differential, then it's wrong to say for example [MATH]f(x) = x^2dx = 2x[/MATH], instead, it is true to say [MATH]df(x) = x^2dx[/MATH] and therefore [MATH]f(x) = 2x[/MATH] like a normal differential equation.
Am I right?

Here's where I found it:
There was this slide about the work of a force that showed how to get the formula [MATH]dW = \frac1 2 md(v^2)[/MATH] where W is the work, m is the mass and v is the speed.
The steps were:
Given that [MATH]dW = F\cdot dr[/MATH],
[MATH]dW = F\cdot dr = F\cdot vdt = m\frac{dv}{dt}\cdot vdt = m\cdot vdv[/MATH] where t is the time;
if [MATH]d(v^2) = 2vdv[/MATH] then [MATH]dW = \frac 1 2 md(v^2)[/MATH]Which didn't make much sense to me at the beginning because [MATH]dW[/MATH] was notated as [MATH]\delta W[/MATH] and I didn't realize that it was just a normal differential.
To conclude, if you divide both sides for [MATH]dp[/MATH] where p is the variable of the position in space, and then integrate both sides for p:
[MATH]\int_{p_1}^{p_2} \frac {dW}{dp}dp = \int_{p_1}^{p_2}\frac 1 2 m\frac {d(v^2)}{dp}dp[/MATH] (yeah, it's just a normal integration but to be complete I also divided by [MATH]dp[/MATH])
i get
[MATH]W(p_2)-W(p_1) = \frac 1 2 m(v(p_2))^2 - \frac 1 2 m(v(p_1))^2[/MATH].

So if I name [MATH]E_k(p) = \frac 1 2 m(v(p))^2[/MATH] the kinetic energy, that equation shows that the work made by a force is the difference of the kinetic energy of the object in the positions:
[MATH]W(p_2)-W(p_1) = E_k(p_2) - E_k(p_1)[/MATH]

 

[Dr.Peterson]

I'm not sure if you have any remaining questions! But as for ...

But I am still not getting the sense of writing something like [MATH]g(x) = f(x)dx[/MATH].
Is [MATH]g(x)[/MATH] equal to [MATH]\int f(x)dx[/MATH]?
Or does it is have a different meaning?

... I would say we wouldn't write [MATH]g(x) = f(x)dx[/MATH] at all. [MATH]g(x)[/MATH] is a function, while [MATH]f(x)dx[/MATH] is a differential, and they can't be equal.

Ok, I've been thinking about it pretty much and maybe it makes sense now.
That just means what I see, a function for the differential of a variable, which actually isn't an integral.
The integration operation "automatically" multiplies the differential, then it's wrong to say for example [MATH]f(x) = x^2dx = 2x[/MATH], instead, it is true to say [MATH]df(x) = x^2dx[/MATH] and therefore [MATH]f(x) = 2x[/MATH] like a normal differential equation.
Am I right?

Yes. The integral is the inverse of the differential, as addition is the inverse of subtraction.

The physics you wrote looks right, overall; someone else may have comments on it. But clearly this presentation assumes you are comfortable with differentials.

 

[Uggohe]

I would say we wouldn't write g(x)=f(x)dx\displaystyle g(x) = f(x)dx at all. g(x)\displaystyle g(x) is a function, while f(x)dx\displaystyle f(x)dx is a differential, and they can't be equal.

Yeah, understood thanks! The [MATH]\delta W[/MATH] misled me and I thought it was a function instead of a differential

 

[JeffM]

Well, thank you, that cleared my ideas a lot.
But I am still not getting the sense of writing something like [MATH]g(x) = f(x)dx[/MATH].
Is [MATH]g(x)[/MATH] equal to [MATH]\int f(x)dx[/MATH]?
Or does it is have a different meaning?

What you wrote is not meaningful at all

You have to have differentials on both sides of the equation to form a differential equation.

[MATH]g((y) \ dy = f(x) \ dx \iff G(y) + c = F(x), \text { where } \dfrac{d}{dy} G(y) = g(y) \text { and } \dfrac{d}{dx} F(x) = f(x).[/MATH]
Notice that you do not get a unique solution. You get a family of solutions differing by a constant. You will need additional information to get a unique solution. Welcome to the world of differential equations.