As an example of something close, N.J. Wildberger worked out Algebraic Calculus, done entirely in $\mathbb Q$, really any field, not necessarily complete. He's changed it around a bit since he came up with the following, a derivation of Cavalieri's Quadrature Formula, which is the usual power rule for integration.
Theorem.
For natural $n$:
$$ \int_0^a x^n dx = \frac{a ^{n+1}}{n+1} $$
Let's credit Archimedes for the $n=2$ case, circa 250 BC. Cavalieri proved it for $n$ up to 9 in 1635.
Instead of a method of integration, Wildberger simply postulates some properties we want integrals to have to model signed area:
Linearity. Areas add and scale in the y direction.
$\displaystyle \int_a^b (f(x)+g(x)) dx = \int_a^b f(x) dx+\int_a^b g(x) dx$
$\displaystyle \int_a^b cf(x) dx = c \int_a^b f(x) dx$
Additivity. Areas add in the x direction.
$\displaystyle \int_a^b f(x) dx +\int_b^c f(x) dx = \int_a^c f(x) dx$
Translation Invariance. We can shift areas in the x direction.
$\displaystyle \int_a^b f(x+c) dx = \int_{a+c}^{b+c} f(x) dx$
Scale Invariance. We can scale areas in the x direction. This is the subtlest one of the bunch, but still pretty straightforward.
$\displaystyle c\int_a^b f(cx) dx = \int_{ac}^{bc} f(x) dx$
Normalization. The unit square has an area of one.
$\displaystyle \int_0^1 1 dx = 1$
Assuming these as axioms we can prove Cavalieri’s Quadrature Formula.
Show
$\displaystyle \int_0^a x^n dx = \dfrac{a ^{n+1}}{n+1} $
Proof.
Define
$\displaystyle A_n = \int_0^1 x^n dx$
Using Normalization we see
$A_0 = 1$
We consider the left side of Cavalieri’s Formula. By Scale Invariance,
$\displaystyle \int_0^a x^n dx = a \int_0^1 (ax)^n dx $
By Linearity,
$\displaystyle a \int_0^1 (ax)^n dx = a^{n+1} \int_0^1 x^n dx =a^{n+1} A_n$
So the integral we’re after is
$\displaystyle \int_0^a x^n dx = a^{n+1} A_n $
Let’s call that the result. We'll use it four or so times below. If we can show $A_n = \dfrac{1}{n+1}$ we’re done.
By Additivity,
$\displaystyle \int_0^1 x^n dx + \int_1^{1+c} x^n dx = \int_0^{1+c} x^n dx$
Applying the result (twice),
$\displaystyle A_n + \int_1^{1+c} x^n dx = (1+c)^{n+1} A_n $
$\displaystyle \int_1^{1+c} x^n dx = A_n((1+c)^{n+1}-1)$
We use the Binomial theorem. The constant $1^{n+1}$ cancels with the $-1$ at the end. We really only need the first term after that.
$ \displaystyle\int_1^{1+c} x^n dx = A_n((n+1)c + { n+1 \choose 2}c^2 + \ldots )$
Note the three dots are not some infinite expression but $c^3$ and higher degree terms of a finite sum.
Let’s do the same integral with Translation Invariance and the Binomial theorem.
$\displaystyle \int_1^{1+c} x^n dx $
$\displaystyle = \int_0^{c} (1+x)^n dx $
$\displaystyle = \int_0^{c} (1+ nx + { n \choose 2} x^2 + \ldots ) dx$
which by Linearity is:
$\displaystyle = \int_0^{c} 1 dx + n \int_0^{c} x dx + \ldots$
Applying the result again,
$ =c^1 A_0 + n c^2 A_1 + \ldots $
$ = c + n A_1 c^2 + \ldots$
Putting it together,
$A_n((n+1)c + { n+1 \choose 2}c^2 + ... ) = c + n A_1 c^2 + \ldots $
These two polynomials in $c$ are equal at every $c$ so their respective coefficients on each power of $c$ must be the same. In particular the linear terms must be equal,
$A_n(n+1) c = c$
$A_n =\dfrac{1}{n+1}$
Substituting into the result, that’s what we needed to conclude
$\displaystyle \int_0^a x^n dx = \dfrac{a ^{n+1}}{n+1} \quad\checkmark$
Integration, no limits or infinite processes required.
Reference: Wildberger, N. J. (2002). "A new proof of Cavalieri's quadrature formula". The American Mathematical Monthly. 109 (9): 843–845.
