# B Integration by substitution

1. Feb 27, 2017

### Conductivity

I have seen the wikipedia's proof which can be found here: https://proofwiki.org/wiki/Integration_by_Substitution
However sometimes, we have problems where you have a $d(x)$ times $f(g(x))$ times g prime of x where we use substitution and it works but the proof didn't prove this condition..

I was wondering if you can prove why it works through infinite sums like for example
$( f(g(x)) g^{'}(x) dx + f(g(x+dx)) g^{'}(x+dx) dx + ..... )$
If I can change that to this
$( f(y) dy + f(y+dy) dy + ....$ where you set y = g(x)
It would statisfy me.. But is it possible to prove it using this?

2. Mar 1, 2017

### Ssnow

Substitution works because the chain rule for derivative works:

$[f(g(x))]'=f'(g(x))g'(x)$

if you integrate both sides, you will obtain

$\int [f(g(x))]' dx=\int f'(g(x))g'(x) dx$

that is the same of $\int f'(g(x))g'(x) dx = f(g(x)) + c$ because the integral is the antiderivation so $\int [f(g(x))]' dx=f(g(x))+c$. This formally can be done substituting in the integral $\int f'(g(x))g'(x) dx$ the function $g(x)=t$ so $g'(x)dx=dt$ (we do a substitution on differentials) differentiating and $\int f'(g(x))g'(x) dx=\int f'(t)dt=f(t)+c$, returning in $g$ we have: $\int f'(g(x))g'(x) dx=f(g(x))+c$.

Ssnow

3. Mar 1, 2017

### Conductivity

Oh, Can I think of it as this?
$[f(g(x))]'=f'(g(x))g'(x)$
This is just df/dx So If I make a variable called u= g(x) and then du/dx = g(x) now flipping this over gives you dx/du = 1/g(x) and we can then multiply it to df/dx to get df/du by chain rule and then we integrate with respect to u.

What bothers me is: Why is dx substitute-able rather than a notation to let us know what we are integrating in respect to?
I know that it refers to something infinitesimal and the integration just means the infinite sums of f(x) multiplied by dx which would make sense in these cases
du/dx = 1/u
u du = 1 dx
$\int u du = \int 1 dx$

Now I can apply the same logic to the substitution rule. I can assign a value u and say dx = du/g'(x) which then substitute it and integrate to get the infinite sum of $\int f(u) du$ But I still prefer the first argument.

Excuse me if I am talking nonsense just getting started at integration

4. Mar 1, 2017

### Ssnow

Yes.

I don't know if can help but the fact is that the presence of derivative inside the integral permit you to substitute ''the differential'', that is the espression $g'(x)dx$ (that you can see as an ''incremental quantity'' ) with another much simpler as $du$. Consequence of this: your integral will be simpler then the first (this is similar to the integration by part rule)

No absolutely, your talk is perfectly reasonable .

Ssnow