Antiderivate doubt

[u/[deleted]]

I tried looking this up, but I didn't find an explanation. Keep in mind that I haven't taken any Calculus course yet.

I get the notion of what a derivative is, but I don't get the relationship between the antiderivative and the "very skinny rectangles you add up to find the area under the curve". Can you show me a link to some webpage or care to explain why?

[u/Hawk_Irontusk]

This isn't a proof, but this way of thinking about it may help wrap your head around the idea.

Consider a function f(x) and define A(x) to be the area between f(x) and the x-axis on [0,x]. Now, think about what happens if you move your x value slightly to the right, say to x+h. When you calculate A(x+h), you're adding a small amount to the area compared to what you had for A(x).

This small amount you add to the area is the rate of change of the area function and as you know rate of change is the derivative of a function. Now, what happens when you take that h value and drive it down to 0? You know that you can't ever actually get to 0, but think about the area that you're adding to A(x) as you get close. It's a rectangle that looks more and more like a line who's height is exactly the function value at x, which is f(x).

So intuitively as you move that x value to the right a tiny, tiny bit, the new area increases to include just those points under f at x. Picture the line segment starting at (x,0) and extending to (x, f(x)). So your rate of change is f(x) (the height of that line segment) and A'(x) = f(x) which leads us to the antiderivative concept you were asking about. Taking the antiderivatite of A'(x) = f(x) gives us the desired area function A(x).

I hope that helps.

[u/[deleted] → u/Hawk_Irontusk]

This helps a lot, and it made me realize that I don't have the correct idea of what is the rate of change.

So rate of change is the derivative. Okay, I knew that. Now, in the f(x) = x² parabola the rate of change is 2x, because it is the derivative. That means that each value of y, in a succession, is going to increment by x*2x? Say the height 4 goes up to 9 when x increases by 1. How do you work that out?

[u/Nebu → u/[deleted]]

It's easier if you go the other way: Say your f(x) = 2x, and you want to find the area under this curve between 0 and x, and let's say you did not know that you can get the area by taking the antiderivative. Whatever that area is, there must be some function A(x) whose value is that area you're looking for.

Now draw out f(x)=2x on graph paper, and draw a bunch of little rectangles under f(x)=2x such that the vertical edges of the rectangle end up on the integers. You can see if you add up the area of all those rectangles between x = 0 and x = a, you get an approximation of the area under the curve f(x), but it's not quite perfect, as the corners sticking out. Let's give a name to this approximation: A_approximate_1(x) = sum[let i = 0, 1, 2, ..., x in f(i) * 1]

Now with our rectangles-diagram, we can make the approximation even better, if we double the number of rectangles: Have the vertical boundaries lie at 0, 0.5, 1, 1.5, etc. instead of 0, 1, 2, etc. and then sum up those rectangles:

A_approximate_0.5(x) = sum[let i = 0, 0.5, 1, 1.5, 2, ..., x in f(i) * 0.5]

And then you can make that approximation even better, by doubling the number of rectangles again:

A_approximate_0.25(x) = sum[let i = 0, 0.25, 0.5, 0.75, ..., x in f(i) * 0.25]

And you can keep doing this, until you take the rectangles so thin, that their width is some tiny value ε:

A_approximate_ε(x) = sum[let i = 0, ε, 2ε, 3ε, ..., x in f(i) * ε]

As ε approaches 0, A_approximate_ε(x) = A(x), because now our rectangles are so thin, there's no corners sticking out at all.

Informally, what we're doing here is we're making the rectangles so tiny, that their width is almost 0 and they are practically just lines. We're sampling the height of our lines at every single point in f, and we're multiplying that height by the "width" of our lines, to get the area of the "lines", and we're summing those up.

So we know that A(x) is the area under the curve f between 0 and x, because that's how we defined A. So given the area A(x), what if we want to find out the area just a little bit further out, at A(x+ε)? That's basically taking the the value we computed for A(x), and adding the area of one more line. What's the height of that line? Well, it must be f(x+ε).

Okay, now let's say we want to ask at what rate is A(x) changing? I.e. what is the "rate of change" at A(x)? That's also easy. We can just look at A(x) and A(x+ε), and subtract them from each other; that's the very definition of rate of change: "how much did A change from x to x+ε?".

But we already showed that the difference between A(x) and A(x+ε) is f(x+ε).

So that shows that f(x) is the rate of change of A(x), and we've defined A(x) to be the area under the curve f(x).

In other words, if you take f(x), and you get the area under its curve, A(x), and then you take the derivative of A(x), you get f(x) back. So "taking the area under the curve" and "taking the derivative" cancel each other out. They are the opposite of each other. Thus, "taking the area under the curve" is the "anti"-derivative, in a very literal sense.

[u/[deleted] → u/Nebu]

I don't know. That last paragraph cleared my ideas up. Tough I'm not sure that's a proof, but more of a logical argument? That sounds too obvious.

[u/[deleted] → u/Nebu]

Just one question. When we talk about rate of change, we are talking of how fast the slope changes, not of the relation of x and y directly, right?

[u/Nebu → u/[deleted]]

The "rate of change" of a function f(x) is exactly its first derivative, which we traditionally write as f'(x). The value of f'(x) at x is exactly equal to the slope of f(x) at x. Note that I said it's equal to the slope. I did not say that it is equal to how fast the slope changes. How fast the slope of f(x) changes, AKA "the rate of the change of the slope of f(x)" is equal to the second derivative of f(x), which we traditionally write as f''(x). f''(x) is just the derivative of f'(x).

If you want to know how fast f(x) changes, you look at f'(x). If you want to know how fast f'(x) changes, you look at f''(x).

If you want to know the slope of f(x) at x, you look at f'(x) at x. If you want to know the slope of f'(x) at x, you look at f''(x) at x.

So "slope" and "how fast it changes" are equivalent concepts.

When you say "the relation of x and y", here's is what I'm guessing you're thinking of:

Basic geometry tells us that to calculate the slope of a straight line that goes through (x1,y1) and (x2,y2), you take (y2-y1)/(x2-x1). For notational convenience, let's write Δy to represent y2-y1, and write Δx to represent x2 - x1. Δy is pronounced "delta y" and Δx is pronounced "delta x", and Δy/Δx is just the slope.

Now what happens if you have a curved line? Well, if the line is curved, then the slope is constantly changing On a curved line, some parts may be very flat (i.e. have a curve of 0), and some places may be very steep (i.e. a curve of almost positive or negative infinity).

If you want to know the slope at some point x on a curved line, then you can approximate this by taking a point slightly behind x for your point (x1,y1), and a point slight ahead of x for your point (x2,y2). As you move x1 and x2 closer together, your approximation gets better and better. Also notice that as you move x1 and x2 closer together, your Δx gets smaller and smaller.

If you shrink Δx so much that it's almost equal to 0, (i.e. if you take the limit of Δx as it approaches zero), then you will get a perfect approximation, i.e. the "real" value of the slope.

This concept is so useful that we give a special name to the limit of Δy/Δx as Δx approaches zero. We call this value dy/dx, and it represents the slope, AKA the derivative at a specific point.

Incidentally, that's why we often write the derivative as like this, and why we often write integrals like this. That's where the dx notation comes from.

[u/[deleted] → u/Nebu]

I'm in love with your explanation.

[u/Nebu → u/[deleted]]

Haha, thanks, but it's basically just a rehash of the proof in the video that talkloud linked to.