I’ve updated @Mistral@lemmings.world to remember the whole tree of comments when replying to you, so it can do follow-up responses. You still have to mention it on every message, though. Example in the comments.
I’ve updated @Mistral@lemmings.world to remember the whole tree of comments when replying to you, so it can do follow-up responses. You still have to mention it on every message, though. Example in the comments.
@Mistral@lemmings.world Please continue.
G(x) = C for all x in [a, b]
Now, let’s evaluate G(a) and G(b):
G(a) = g(a) - h(a) = g(a) - [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
G(b) = g(b) - h(b) = g(b) - [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Subtracting G(a) from G(b), we get:
G(b) - G(a) = g(b) - g(a) - [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] + [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Simplifying this expression, we get:
g(b) - g(a) = [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
This is exactly what we wanted to show. Therefore, we have derived the Fundamental Theorem of Calculus, Part 2:
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = g(b) - g(a)
This theorem states that if a function f(x) has an antiderivative g(x), then the definite integral of f(x) over the interval [a, b] is equal to g(b) - g(a), where g’(x) = f(x).
@Mistral@lemmings.world I think the proof of the second part was fine; it was the first part where the interchange of limits was used and was therefore erroneous.
You are correct. My apologies for the confusion. The first part of the Fundamental Theorem of Calculus should be proved using the concept of Riemann sums, which avoids the interchange of limits issue.
Here’s the corrected proof for the first part of the Fundamental Theorem of Calculus:
Let f(x) be a continuous function on the interval [a, b]. We want to show that f(x) has an antiderivative g(x) such that g’(x) = f(x).
Define a function G(x) as follows:
G(x) = ∫(a, x) f(t) dt
Here, the integral is taken over the interval [a, x]. We want to show that G(x) is an antiderivative of f(x), i.e., G’(x) = f(x).
Now, consider a partition P of the interval [a, b]:
P: a = x0 < x1 < … < xn = b
Let x_i* be any point in the interval [x_(i-1), x_i]. We can construct a Riemann sum for the integral:
S(P, {x_i*}) = Σ (from i = 1 to n) f(x_i*) * (x_i - x_(i-1))
By the definition of the integral, as the maximum width of the intervals in the partition P approaches 0, the Riemann sum S(P, {x_i*}) approaches the definite integral:
lim (max(P) -> 0) S(P, {x_i*}) = ∫(a, b) f(t) dt
Now, let’s construct a sequence of partitions {P_k} such that max(P_k) -> 0 as k -> ∞. For each partition P_k, let’s choose a point x_k* in the interval [x_(k-1), x_k]. Then, we have:
G(x_k) - G(x_(k-1)) = ∫(x_(k-1), x_k) f(t) dt = S(
For anyone else reading this [1], a partition of an interval [a,b] is a finite set of points satisfying the condition a = x_0 ≤ x_1 ≤ … ≤ x_n-1 ≤ x_n = b. So the max(P) = b, so the condition max(P) -> 0 just means b -> 0, which is wrong.
[1] But like, y tho?
@Mistral@lemmings.world In the previous output, I think that you mean “max(norm(P))” where you say max(P), right? The condition “max(P) -> 0” just smooshes the partitions into the origin.
You are correct. I apologize for the confusion. The correct statement should be:
As max(norm(P)) -> 0, the Riemann sum S(P, {x_i*}) approaches the definite integral:
lim (max(norm(P)) -> 0) S(P, {x_i*}) = ∫(a, b) f(t) dt
@Mistral@lemmings.world Okay. Now can you rewrite the proof in language a five-year-old would understand?
Of course! Let’s imagine a long pathway, like a hallway, that goes from a door at one end (point A) to another door at the other end (point B). We want to find out the total distance of this hallway.
Now, let’s pretend we don’t know the exact distance, but we can find the distance between some points along the hallway. We can do this by placing markers on the floor, like stickers, and measuring the distance between them. We can call these markers “partitions” and the distance between them “widths.”
When we make these partitions closer and closer to each other, our estimate of the total distance of the hallway will get better and better. In the end, we want to find out the exact total distance of the hallway, which is what we call the “definite integral.”
So, imagine we have a friend who can tell us the height of the hallway at any point. We can think of the height as the “function” we want to integrate. Now, we can make a “Riemann sum” by adding up the height of the hallway at each partition and multiply it by the width of that partition.
As we make more and more partitions and the widths of the partitions get smaller and smaller, our Riemann sum will get closer and closer to the actual total distance of the hallway, which is the definite integral we’re trying to find. And that’s how we can find the definite integral by using partitions and Riemann sums!
@Mistral@lemmings.world That’s great! Can you rewrite the proof so a dog can understand it?