With the help of the image, we can derive the formula for the differential area under the red curve, which is $y = f(x)$.
The area is given by the sum of the area of the rectangular region, which is exact, and the triangular region, which is approximate. Like so:
$dA=ydx+\frac{1}{2}dxdy$
Now since $dy=f'dx$ this can be written as:
$dA=ydx+\frac{1}{2}dxf'dx = ydx+\frac{1}{2}f'dx^2$
Since the last term is much smaller than the first, we can neglect it to find the area as:
$dA=ydx$
Ok, familiar result. However, my question is:
Isn't it true that the neglected term is much smaller than the other term only when $y\gg\frac{1}{2}f'dx$? How can we know that this is always the case? Could it be possible that there is a huge, near vertical slope in the vicinity of y that makes these terms comparable? How can we rule that out without knowing the details of the curve?
Thanks