# Integral physics, please help me understand a thing with respect to integration

1. Nov 11, 2014

### christian0710

Hi, i'm trying to understand why

When you write a*dt = dv then you can write the integral like this.,

∫dv (from v0 to vt) = ∫a*dt (from 0 to t)

My challenge is this: from the equation a*dt = dv, the term "dv" geometrically means an infenitesimalle small change in function value of the function v(t), so dv must be an integer. So you are integrating an integer, dv, and the graph of an integer is a straight horizontal line on the v(t) vs t axis. so is it correctly understood that if we assume dv= 1 then the integral of dv should just be v. But here is the part i don't understand, the lower and upper bounds are v0 and v, usually the limits of integration are limits on the x-axis and Not on the Y-axis (or V(t) axis), so should i interprete dv as an integer whose function has Velocity on the x axis? This just does not make sense to me,

Here is a photo of what is written in the book.,

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2. Nov 11, 2014

### mathman

Absolutely no.

All the equation is saying is that the change in velocity is the integral of the acceleration.

3. Nov 11, 2014

### Staff: Mentor

Why do you think that?

Are you perhaps confusing the words "integer" and "constant"?

4. Jan 17, 2015

### Svein

dv = a*dt just means that an acceleration creates a change in velocity. The amount of change can be calculated using integration (or in the simple case, where the acceleration is constant, multiplying the acceleration by the time passed).

5. Jan 17, 2015

### PWiz

The equation can be rearranged to give you $a=\frac{dv}{dt}$ , and dividing two infinitely small numbers results in a definite one($a$ over here). Note that both $dv$ and $dt$ mean changes in the respective quantities over an infinitesimally small interval, and this in no way suggests that they always equal to an integer. It appears that you've misunderstood the concept of the $dx$ terminology, so I recommend that you go through this article: http://www.felderbooks.com/papers/dx.html

6. Jan 18, 2015

### Svein

I always thought that "infinitesimals" concept died out around 1880...

7. Jan 18, 2015

### PWiz

With the $dx$ notation, yes.

8. Jan 18, 2015

### SteamKing

Staff Emeritus
Maybe for certain schools of pure mathematicians, but the infinitesimal concept was revived in the TwenCen to serve as a basis for calculus and analysis.

See:

http://en.wikipedia.org/wiki/Infinitesimal

As you might have noticed, infinitesimals still heavily populate texts on physics and calculus, among other subjects.

9. Jan 18, 2015

### Svein

Well, having the equivalent of a Master's in complex analysis, that makes me a "pure mathematician", I suppose.

10. Jan 18, 2015

### PWiz

Well I haven't even finished high school yet, makes me feel a tad bit insecure haha