# Change of variable in indefinite integrals

1. Oct 28, 2006

### quasar987

I've had this very basic question on the back of my mind for 2 almost years now and I think I've found a satisfactory answer. The question is this most simple one: how do we justify a change of variables such as

$$\int e^{ax}dx = \int \frac{1}{a}e^udu$$

in an indefinite integral? My "solution" is this... We are looking for a primitive $F(\beta)$ to $f(\beta)$. By one version of the fondamental thm, as soon as f is continuous, such a primitive is given by

$$F(\beta)=\int_{\phi(\alpha)}^{\phi(\beta)}f(x)dx = \int_{\alpha}^{\beta}(f\circ \phi)(s)\frac{d\phi}{ds}ds$$

where $\phi:(\alpha,\beta)\rightarrow (\phi(\alpha), \phi(\beta))$ is $C^1$. (Ths is just the change of variable thm)

This may be total nonsense, but my eyes are itching from tiredness so I wouldn't know. Any comments on the above or an opinion on how to justify change of variable in indefinite integrals will be warmly appreciate tomorrow morning.

Good night!

2. Oct 28, 2006

### HallsofIvy

Staff Emeritus
I'm not at all sure what you are asking. Certainly if
$$F(x)= \int e^{ax}dx$$ and
$$G(u)= \frac{1}{a}\int e^u du$$
then, by the chain rule, with u(x)= ax,
$$\frac{dG}{dx}= \frac{dG}{du}\frac{du}{dx}$$
$$= \left(\frac{1}{a}e^u\right)(a)$$
$$= e^u= e^{ax}= \frac{dF}{dx}$$
Since F and G have the same derivatives they are anti-derivatives (indefinite integrals) of the same thing.

"Substitution" in integrals is just the chain rule "in reverse".

3. Oct 28, 2006

### quasar987

ack, I had tried things with the chain rule, but never this. Yep, this is exactly what I was loking for, thanks Halls!