So I get the idea that the area underneath a part of a graph is the same as the sun of all the infinitely small rectangles that make it up. I also get that you can find the area by between two points by plugging in two values into the equation for the anti differential and finding the difference. I just don’t understand why. What has the anti differential even got to do with anything? Where did it come from and how does it relate to the rectangles? I’ve tried finding other explanations online and on elsewhere on Reddit but it still doesn’t make much sense to me.
Edit: I meant anti derivative not anti differential lol
In: 1
Let’s think about this in terms of d/dx of integral from 0 to x f(t) dt. We want to show that this is, in fact, f(x).
What does this mean? Well, d/dx is how much the value changes as *x* changes by some small amount *dx*. In other words, it’s the difference between integral from 0 to x of f(t) dt and 0 to x+dx of f(t) dt, where we think of *dx* as an infinitely small, non-zero number, divided by dx. In other words, our goal is to calculate:
* (integral 0 to x+dx f(t) dt) minus (integral 0 to x of f(t) dt), all divided by dx.
The basic rules of calculus tell us that integral from a to c of f(t) dt is just integral from a to b of f(t) dt plus integral from b to c of f(t) dt. Intuitively, we’ve just split the area under the curve into two regions. So we can write integral 0 to x+dx f(t) dt as a sum of two integrals: integral from 0 to x of f(t) dt (the original value) plus the integral from x to x+dx of f(t) dt (the “change” as x increases to x+dx). Then we can substitute this into the expression above:
* (integral 0 to x f(t) dt plus integral x to x+dx f(t) dt) minus (integral 0 to x f(t) dt), all divided by dx
The first and third values in the numerator cancel out: they’re both integral 0 to x f(t) dt. So we’re left with:
* integral x to x+dx f(t) dt, divided by dx
So what the heck is integral x to x+dx f(t) dt?
Well, since dx is very small, f(t) is effectively constant on this interval. In fact, since we’re thinking of dx as infinitely small, f(t) is **exactly** constant on this interval. And like all integrals of a constant, the area under the curve is just the width of the integral times the height. The width is from x to x+dx, which is just x+dx-x = dx. And the height is just f(x), the actual height of the curve we’re integrating at this value. So integral x to x+dx of f(t) dt is just f(x) dx. And we can substitute:
* (f(x) dx) divded by dx
* = f(x)
Which is exactly what we hoped to prove. [Here’s a diagram to show what we just did algebraically](https://i.imgur.com/TdjaQ6K.png).
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