Is there a proof that the area under a curve is equivalent to the definite integral, that doesn't involve the fundamental theorem of calculus. Perhaps a proof that uses Riemann sums.
[Math] Proof that the area under a curve is the definite integral, without the fundamental theorem of calculus
integrationriemann sum
Best Answer
I feel your question is a little bit reversed in terms of the logical sequence of integration and area. So firstly the question arises about how we calculate the "area" beneath a function, and then people invented integration to do that, and people found that the definition of the "area"/measure is tricky.
Riemann integration tries to give strict definition of how to define the "area" under a function by vertically partition it with small rectangles to approximate the "area" and take the limit of the sum. But it has limitations in dealing with function limits (e.g. Fourier Series of a function) and also some functions do not have Riemann Integration ( e.g. function that is 1 for irrational numbers in $(0,1)$, and 0 otherwise)
Lebesgue integration solved this issues by firstly defining measure over sets, and then add up horizontally by multiplying function values with the measure over the sets that is mapped around the value by the function. It is compatible with Riemann integration. The Lesbegue monotone convergence theorem etc. provides good tools for analyzing the integration of limit functions, and actually the Lebesgue integration is firstly defined on simple functions, and then extend to non-negative function using those limit theorem, and then extend by $f=f_+-f_-$. Also some function that is not Riemann integrable is Lebesgue integrable.
In sum, the integration arises when people want to calculate the area under the function, but then people realized the definition of "area"/measure is actually very tricky.