Locally Flat

Assume we want to calculate the area under

y = cos(x)

from

x = 0 to x = \dfrac {\pi} {2} (we are using radians)

We divide the interval from x = 0 to x = \dfrac {\pi} {2} into equally spaced segments and put rectangles on the segments. The height of a rectangle is the value of the function on the right side of the rectangle. The top edge of the rectangle will always be under the function because, going from right to left, the function is always increasing.

We know that as the number of rectangles gets larger and larger, the difference between the area under the function and the sum of the area of the rectangles approaches zero. In fact we can get that difference is close to zero as we want by making the number of rectangles larger and larger. Knowing this, we use an idea from the study of limits to say that we know that as the number of rectangles becomes infinite the area of the sum of rectangles equal the area under the function.

We also know that it wouldn’t be possible for the rectangle summation to equal the function area unless for each interval, the rectangle area equaled the function segment area– this could only happen if over the span of that interval, the function was flat.

Putting all this together we say the function is flat over a local region with width dx.