$ \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 $a<b$. 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.

  • $\begingroup$ I really like this one because it doesn't require the multiplication property of dx. $\endgroup$ – GLaDOS Mar 29 at 11:28
  • $\begingroup$ But can you prove it for indefinite integral? $\endgroup$ – GLaDOS Mar 29 at 11:44
  • 1
    $\begingroup$ @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, $$\int_a^b 3x^2\sin x^3 dx=\int_{a^3}^{b^3}\sin udu\implies\int 3x^2\sin x^3 dx=\int\sin udu.$$ $\endgroup$ – J.G. Mar 29 at 12:02
  • 1
    $\begingroup$ @GLaDOS That's a special case of the fundamental theorem of calculus, viz. $f(b)=\frac{d}{db}\int_a^b f(x)dx$. $\endgroup$ – J.G. Apr 18 at 19:55
  • 1
    $\begingroup$ oh ok thanks. That actually makes sense. $\endgroup$ – GLaDOS Apr 18 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.