Can Derivative Operators Be Treated Like Fractions in Integration?

[ronaldor9]

How is it possible that one can break up the derivative operator such as this:

\frac{dv}{dt}=t^2, then integrate like this,
\int^v_{v_{0}}dv = \int^t_0 t^2 dt, wherev=v_{o} when t=0. Especially in light of what most calculus teachers tell you; that the derivative symbol is not a fraction and should not be interpreted as a faction?

 

[Count Iblis]

The derivative is a limit of a fraction and the integral is a limit of a summation of function values times a step length.

 

[slider142]

dv/dt is a function v'(t), and since it is a derivative, it is integrable by the fundamental theorem of calculus to v(t) + C where C is an undetermined constant.
If f(t) = g(t) and f(t) is integrable, then \int f = (\int g) + C where C is an undetermined constant. Your use of boundary conditions (v(0) = v₀) allows you to determine the constant.
That's all that's being done here. The Leibnitz notation is just a neat way of writing it out, but don't take it too seriously until you learn the proper way to manipulate differential forms.

 

[ronaldor9]

wow thanks slider. I never have thought about it that way, but now that you have explained it, it is very interesting!

 

[Random Variable]

This was discussed in another thread.

What you're really integrating on the right side is \int^v_{v_{0}} \frac {dv}{dt} dt