# Integral and a derivative definition

1. Mar 22, 2013

### 71GA

Hello,

I just whant to know what mathematical rule alows me to do this? I mean i think it is u-substitution, but i am not sure how it is done here? It is weird to me as it seems that $dt$ just cancel out and limits are changed...

$$\int\limits_{0}^{t} \frac{dv}{dt} \cdot mv \gamma(v)\, dt = \int\limits_{0}^{v} mv \gamma(v)\, dv$$

2. Mar 22, 2013

### pwsnafu

Integration by Substitution:
Suppose $I$ is an interval. Let $g : [a,b] \to I$ have continuous derivative. Let $f : I \to \mathbb{R}$ be continuous. Then
$\int_{g(a)}^{g(b)} f(x) \, dx = \int_{a}^{b} (f \circ g)(t) \, g'(t) \, dt$

3. Mar 22, 2013

### 71GA

This is a basic formula yes i know, but how does this solve my case?

4. Mar 22, 2013

### HallsofIvy

"Integration by substitution" is, essentially, the "chain rule" reversed. It is valid because the chain rule for differentiation is valid.

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