Csdrhrt

I am trying to derive the area under a curve in the cartesian plane. To begin I noted the values of the curve at x and $x+Delta x$ (as $y$ and $y+Delta y$ respectively), found their average, and multiplied this by the distance $Delta x$, effectively giving a the area of a rectangle of width $Delta x$ and height $frac{y+Delta y}{2}$ This gave the area for this segment as:

$$
A = frac{y +(y+Delta y)}{2} Delta x = y Delta x + frac{Delta y Delta x}{2}
$$

Summing an increasing number of these rectangles over an interval as $Delta x$ goes to zero (I'm not familiar with the proper language) would, I believed, give the correct answer if I only had the first term, and not the second, of the second equation.

I believe I am going wrong by separating $y+Delta y$ into y and $Delta y$. If I am not, does the second term disappear in the limit, and if so, why should the second but not the first disappear?

Using a 'functional' approach:

$$
A = bigg(frac{f(x+Delta x)+f(x)}{2}bigg) Delta x
$$

I recognise the fraction inside the brackets as being nearly the derivative from first principles but don't see how I can find the limit. Can I take the limit of the entire thing as $Delta x$ goes to zero to get the integral for the area under a curve? How would I do this? Thank you

What you are trying to do is to find the area under a positive function
over an interval $[a,b]$ using the midpoint quadrature.

First you divide the interval $[a,b]$ into $n$ equal parts and let $Delta x = frac {b-a}{n}.$

You need to find the sum $$sum _{i=0}^{n-1} f(m_i) Delta x$$ where $$m_i= x_i +frac {Delta x}{2}$$
Then you find the limit of the sum as $nto infty $

Note it has been almost 30 years since I last studied, used or taught calculus, so I'm rather rusty. Thus, my response may not be as precise & correctly worded as others can make it, but I hope it helps.

In answer to your first question, the second term "disappears" in the limit because as $Delta x$ goes to 0, so does $Delta y$. This means that the second term becomes infinitesimally small compared to the first term, so its relative contribution to the sum goes to 0.

Also, although taking the average of the 2 heights, i.e., $y$ and $Delta y$, is more accurate than just using the value of $y$ for any given $Delta x$ for determining the area of the rectangle, the error is the difference in the 2 areas is $cfrac{y + (y + Delta y)}{2} Delta x - y Delta x$ which simplifies to your second term, i.e., $cfrac{Delta x Delta y}{2}$. Thus, as mentioned above, it's not really necessary as the limit of this will go to 0 and, as such, it basically complicates the mathematics unnecessarily.