Why is area under curve derived using $y\,\mathrm dx$ instead of $y\,\mathrm dl\;?$ We know that while deriving the surface area of cone, for differential area, we multiply the slant height element and the circumference of a small circular portion.
So when we calculate the area under curve, why don't we take $y\,\mathrm dl$ where $\mathrm dl=\mathrm dx\,\sec\alpha,$ where $\tan\alpha$ is the instanteneous slope at that $(x,y)\;?$ I think this is more appropriate since the error is not like in the case of assuming $y\,\mathrm dx$ with rectangles where some portion of area is unaccounted for.


Also known as a Riemann sum, the left hand rule version of a definite integral is
$$\int_a^b f(x)dx=\lim_{N\to\infty}\frac{b-a}N\left(f(a)+ f\left(a+\frac{b-a}N\right)+ f\left(a+2\frac{b-a}N\right)+… f\left(a+(N-1)\frac{b-a}N\right)\right)=\lim_{N\to\infty}\color{darkgray}{\frac{b-a}N\sum_{n=0}^{N-1}f\left(a+n\frac{b-a}N\right)}$$
while the formula for the trapezoid rule form of one is:
$$\int_a^b f(x)dx=\lim_{N\to\infty}\frac{b-a}{2N}\left(f(a)+2f\left(a+\frac{b-a}N\right)+2f\left(a+2\frac{b-a}N\right)+…+2f\left(a+(N+1)\frac{b-a}N\right) +f(b)\right)=\lim_{N\to\infty}\color{darkgray}{\frac{b-a}N}\left(\frac{f(a)+f(b)}2+\color{darkgray}{\sum_{n=1}^{N-1}f\left(a+n\frac{b-a}N\right)}\right)$$
where $N$ is the number of subintervals used to estimate the definite integral, like in the question’s drawing. Now compare both dark gray parts to see that the trapezoid rule is very similar to the left hand, or right hand or midpoint, rule.
