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
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).
The connection comes from the definition of derivative…it’s usually just referred to as “the slope”, but it’s really, “If I take successively tinier width slices, what value does the average slope converge to?”. Average slope is “height at one edge minus height at the other edge divided by width.” And the area of *that* shape (a tall thin “rectangle” with a sloped top) has *exactly* the same area as an actual rectangle who’s height is the average of the left and right sides (you can doodle on paper to visualize it).
So adding up a bunch of tiny rectangles whose height is the curve is really the same as adding up a bunch of tiny “sloped top slices” that hug the actual curve. And the integral (anti-derivative) is, mathematically, the area of adding up all those little sloped-top slices.
The connection between anti-derivatives and areas is exactly the content of the [Fundamental Theorem of Calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus). DO you want an explanation why that is true? Or something else? It is however a theorem, not by definition.
Let’s use an example.
We understand that the derivative of a graph of distance over time gives you velocity over time.
Therefore the anti-derivative of a graph of velocity over time should give you back your distance over time.
So now we’ve transferred the question: why does the area under a graph of velocity over time give you the distance?
Well, easy. Velocity is distance over time. So multiplying that by time gives you back distance as the time and “over time” cancel out (e.g. m/s x s = m).
If you had a graph of constant velocity, then it would be abundantly clear that velocity x time corresponds to the rectangle under the graph, with velocity being the lengths of the vertical sides of that rectangle and time being the length of the horizontal sides.
When the graph is curvy you have to break it up into infinite, infinitesimally small rectangles to get that area, but the area still represents the distance (e.g. the antiderivative) nonetheless.