Why is the result of the integral the entire area under the function?
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You can think of an integral as having two important pieces.
1) On it’s own it’s an infinitely thin slice of the area below a function, a magnitude really. A value with no side-to-side measurement (because it’s infinitely thin) and purely a vertical measurement, a ‘height’.
2) The assumption is you are defining a starting point and an ending point for the integration. So you are *integrating* (adding together) a infinite number of infinitely thin slices from the start to the finish. Now the widths *do* add up to something measurable *and* you have all the heights already. In a sense you now have a shape and you can width x height to get the area of the shape. That’s essentially what you are actually doing. Instead of calculating a base x height area for an easy rectangle, you are calculating a complicated base x height for a wonky shape.
Because the definition of the relationship between a derivative and its integral is the tangent!!
i.e. The “Tangent” (rise/run) is the SIMPLEST way for f(x) to change; a straight line from here to over yonder. We could instead consider; does this thing change like a sine wave but then that wouldn’t be comparing things to changing like “straight lines” it’d be comparing to sines!
So, if something is perfectly “line-like” then f'(x)=m and f(x)=mx+b. Eval’ed at boundaries a & b results in… (b-a)[f(b)-f(a)]/2 …which is literally the area of a right angle triangle with base = (b-a) and height [f(b)-f(a)]
So, the definition only yields areas under curves when you use *a particular calculus*, the one often referred to as The calculus. That is, one where we’re comparing things changing like simple straight lines.
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