>>11782861Going through math "blindly" has given you a setback on the ability to intuitively understand math. Everyone has different methods of intuition so it can't really be in a textbook or be particularly emphasized in a lesson, it's just a skill that you have to learn independently over time.
Without reading the next section, try to figure out intuitively why the area of a trapezoid = (big base + little base) * height / 2.
The way I think of it is that if you scan over the trapezoid from top to bottom measuring the area, you're gaining area at a rate of the width per unit of height (in a 2x5 rectangle, you scan over it and gain 2 units of area per unit of height, over 5 units of height, for 2 * 5 = 10 units of area), but the trapezoid's width changes, so intuitively, I realize that (big base + little base) / 2 is the average width, so that just gets multiplied by the height for the area. If you squish a trapezoid into a rectangle of the same height and area, the width of the rectangle is the trapezoid's average width. Since the width of the trapezoid varies linearly, the average can be unweighted.
This next one is for if you've taken Calc I:
Try to think of why the product rule holds true. The product rule states that if f(x) = g(x)h(x) then (sorry for mixed notation, f' needs kerning) d/dx f(x) = g'(x)h(x) + g(x)h'(x).
Here's my geometric intuition for it. Imagine that f(x) is the area of a rectangle whose width is g(x) and whose height is h(x), with bottom left corner at (0, 0) and the upper and right sides moving around as x varies. The width varies at a rate of g'(x) and the height varies at a rate of h'(x). The portion of d/dx f(x) coming from the right side of the rectangle is equal to the rate of change of its width, times its height, so g'(x)h(x). Likewise, the portion of d/dx f(x) coming from the upper side of the rectangle is equal to the rate of change of its height, times its width, so g(x)h'(x) (order has been switched to match the formula).
