# What is the rigor behind U subsitution?

$$\int f(g(x)) dx = \int \frac {f(u)}{u'} du$$

requires that

$$\int f(x) dx = \int f(x) \cdot dx$$

but dx just represents the variable that F(x) +c is a function of. So why is it legal for dx to be treated algebriacally?

I tried investigating this property by using riemann summation:

$$\lim\limits_{n \to \infty}(\sum_{k=1}^n f(\frac{kx}{n} )\frac{x}{n}) = \int_0^x f(t)dt= (F(x)+c)-(F(0)+c)$$

and so you can define

$$\lim\limits_{n \to \infty} (\sum_{k=1}^n f(\frac{kx}{n} )\frac{x}{n})+F(0)+c) = \int f(x) dx$$

you can write $$\frac{x}{n} = dx$$ and $$k\frac{x}{n}=kdx= x_k$$

then you have

$$\lim\limits_{n \to \infty} (\sum_{k=1}^n f(x_k)dx)+F(0)+c) = \int f(x) dx$$

since dx is being written to be multiplied by the series, then you can define dx in different terms to aquire an integral in terms of other variables.

but that exists only in abstraction. I can't quite sufficiently complete the task of doing so.

Let's prove that $$\int_a^b h(g(x)) g^\prime(x) dx=\int_{g(a)}^{g(b)} h(u) du$$ for $$g$$ monotonic on $$[a,\,b]$$ with $$a. Let $$H$$ denote an antiderivative of $$h$$, without loss of generality satisfying $$H(a)=0$$. Then the right-hand side of the putative result is $$H(g(b))-H(g(a))$$. Differentiating this with respect to $$b$$ gives $$h(g(b)) g^\prime(b)$$, so $$H(g(b))-H(g(a))=\int_a^b h(g(x)) g^\prime(x) dx$$ as required.
• @GLaDOS An indefinite integral just means "your favourite antiderivative $+C$", so if substitution works for definite integrals it also does for indefinite ones. For example, $$\int_a^b 3x^2\sin x^3 dx=\int_{a^3}^{b^3}\sin udu\implies\int 3x^2\sin x^3 dx=\int\sin udu.$$ – J.G. Mar 29 at 12:02
• @GLaDOS That's a special case of the fundamental theorem of calculus, viz. $f(b)=\frac{d}{db}\int_a^b f(x)dx$. – J.G. Apr 18 at 19:55