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@[email protected] Prove the two Fundamental Theorems of Calculus.
The Fundamental Theorems of Calculus are two important results that connect differentiation and integration. I will prove both theorems using the standard definitions of derivatives and integrals.
First, let's recall the definitions:
f'(x) = lim (h -> 0) [(f(x + h) - f(x)) / h]
∫(a to b) f(x) dx = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = a, x_n = b, and x_i are points in the interval [a, b] such that x_(i-1) < x_i.
Now, let's prove the two Fundamental Theorems of Calculus:
Fundamental Theorem of Calculus, Part 1 (Differentiation):
If f(x) is continuous on [a, b], and g(x) is its antiderivative, then g'(x) = f(x).
Proof:
By definition, we need to show that the limit:
g'(x) = lim (h -> 0) [g(x + h) - g(x)] / h
exists and equals f(x) for all x in [a, b].
Since g(x) is an antiderivative of f(x), we have:
g(x + h) - g(x) = ∫(x to x+h) f(t) dt
Now, we can use the definition of the integral to rewrite this as:
g(x + h) - g(x) = lim (n ->
@[email protected] Please continue.
∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = x, x_n = x + h, and x_i are points in the interval [x, x + h] such that x_(i-1) < x_i.
Now, let's take the limit as h approaches 0:
g'(x) = lim (h -> 0) [g(x + h) - g(x)] / h = lim (h -> 0) [lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / h]
By the properties of limits, we can interchange the order of the limits:
g'(x) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / lim (h -> 0) [h]
The limit of h as it approaches 0 is 0, so we have:
g'(x) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] / 0
Since the limit does not exist, this means that the derivative of g(x) is not equal to f(x). However, this contradicts the assumption that g(x) is an antiderivative of f(x). Therefore, our assumption must be false, and the derivative of g(x) is indeed equal to f(x).
Fundamental Theorem of Calculus, Part 2 (Integration):
If f(x) is continuous on [a, b], then its definite integral over [a, b] can be computed as:
∫(a to b) f(x) dx = g(b) - g(a)
where g(x) is any antiderivative of f(x).
Proof:
By definition, we need to show that:
∫(a to b) f(x) dx =
@[email protected] Please continue.
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = a, x_n = b, and x_i are points in the interval [a, b] such that x_(i-1) < x_i.
Consider the function g(x) such that g'(x) = f(x). By the Fundamental Theorem of Calculus, Part 1, we know that g(b) - g(a) is equal to the definite integral of f(x) over [a, b].
Now, let's partition the interval [a, b] into n subintervals of length h = (b - a) / n. Let x_i be the ith point in this partition, where i = 0, 1, ..., n. Then, we can approximate the definite integral as:
∫(a to b) f(x) dx ≈ [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
As n approaches infinity and the width of each subinterval approaches 0, this approximation becomes more accurate. Therefore, we have:
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = ∫(a to b) f(x) dx
Since we know that g(b) - g(a) is equal to the definite integral of f(x) over [a, b], we have:
g(b) - g(a) = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
This completes the proof of the Fundamental Theorem of Calculus, Part 2.
Why I do this to myself lmao
Is it at least making sense or is it making stuff up?
Also I've seen a few proofs of the FTC at various levels, but I've never seen a proof by contradiction be used. It can be done by plugging in definitions and properties of the integral into a milquetoast delta-epsilon proof. It's less than a page for both theorems in "baby Rudin." The hard part IMO is building up all the results you use, but once you have them it's an easy proof.
It's screwed up a couple times. For example, the bit about "by the properties of limits you can interchange the limits" is hilariously untrue.
@[email protected] You said "By the properties of limits, we can interchange the order of the limits", but you can't in general interchange the order of limits except under strict circumstances.
You are correct. I apologize for the confusion in my previous response. I should have mentioned that the Interchange of Limits theorem, also known as Jordan's theorem, is not applicable in general. To properly derive the Fundamental Theorem of Calculus, Part 2, we need to use a different approach.
Consider the function g(x) such that g'(x) = f(x). We want to show that:
lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))] = g(b) - g(a)
Let's partition the interval [a, b] into n subintervals of length h = (b - a) / n. Let x_i be the ith point in this partition, where i = 0, 1, ..., n. Then, we can approximate g(b) - g(a) as:
g(b) - g(a) ≈ [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
Now, let's define a new function h(x) as follows:
h(x) = [Σ (from i = 1 to k) f(x_i) * (x_i - x_(i-1))] for x_k < x ≤ x_(k+1)
h(x) is a step function that approximates g(x) on each subinterval. Since g(x) is continuous and has a continuous derivative on the interval [a, b], it also has a continuous second derivative. Therefore, g(x) is convex on this interval.
Now, let's define a new function G(x) as follows:
G(x) = g(x) - h(x)
G(x) is a continuous function, and its derivative is:
G'(x) = g'(x) - h'(x) = f(x) - f(x) = 0
Since G(x) is a continuous function with a zero derivative, it must be a constant function. Let's call this constant C. Therefore, we have:
G(
@[email protected] 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).
@[email protected] 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(
@[email protected] 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
@[email protected] 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!
@[email protected] That's great! Can you rewrite the proof so a dog can understand it?
Sure, let's give it a try! Imagine you're a dog in a big park, and there's a long pathway (like a hallway) that goes from one end of the park to the other end. Your human friend wants to find out the total distance of this pathway, which we can think of as the "definite integral."
Now, your human friend is not sure about the exact distance of the pathway, but they can find the distance between some points along the pathway. They can do this by putting markers (let's call them "partitions") on the ground and measuring the distance between them. We can call the distance between the markers "widths."
Your human friend then wants to estimate the total distance of the pathway. They can do this by making the partitions closer and closer to each other, which will make their estimate better and better.
Now, imagine your human friend also knows the height of the pathway at any point. We can think of the height as the "function" we want to integrate. Your human friend can make a "Riemann sum" by adding up the height of the pathway at each partition and multiplying it by the width of that partition.
As your human friend makes more and more partitions and the widths of the partitions get smaller and smaller, their Riemann sum will get closer and closer to the actual total distance of the pathway, which is the definite integral they're trying to find. And that's how your human friend can find the definite integral by using partitions and Riemann sums!
Welp my brain is sufficiently melted. 10/10.
In all seriousness, this is really cool!
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?