First principles derivation of area under a curve giving rise to an unexpected term before taking limits 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
 A: 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 $n\to \infty $ 
A: 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 just complicates the mathematics unnecessarily.
