The Student Room Group

Fundamental theorem of calculus! So confused...

I have read a few websites and watched a few videos, but for the most part, I am still completely lost. From what I can make sense of. The theorem describes how differentiation and integration are inverses of each other.

If function G(x)= integral^x, a f(t) dx, then the definite integral is g'(x)= f(x)... So I understand why that works based on how the integration works, but I don't understand what it means.

Could someone dumb this down for me, please?

edit: I know there are two parts to the theorem, and there are conflicting arguments as to which is the first and which is the second. I don't know any of it lol
(edited 1 month ago)
Reply 1
Original post by KingRich
I have read a few websites and watched a few videos, but for the most part, I am still completely lost. From what I can make sense of. The theorem describes how differentiation and integration are inverses of each other.
If function G(x)= integral^x, a f(t) dx, then the definite integral is g'(x)= f(x)... So I understand why that works based on how the integration works, but I don't understand what it means.
Could someone dumb this down for me, please?
edit: I know there are two parts to the theorem, and there are conflicting arguments as to which is the first and which is the second. I don't know any of it lol

https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus
is a decent description. So if you have F(x) which is the definite integral of f(t) from a to x, then F'(x) = f(x). Similarly if F(x) is any antiderivative of f(x), then the definite integral of f(x) between a and b is F(b)-F(a). Its important in that it clearly describes how integration and differentiation are related (inverse operations), which may seem reasonably obvious but wasnt so at the time (newton/leibniz)
Another way of thinking about your version of FTC (I think this is the 3blue1brown Essence of Calculus explanation):

Remember, if you fix the value for x, G(x)=axf(t)dtG(x) = \int^x_a f(t) dt is the area under the curve from a to x (trust me on this for now, but I assume you'd agree), and ddxG(x)\frac{d}{dx}G(x) is the change in G(x) given a small change in x (in the picture, the small change is h - this is what d/dx means by definition in words).

rect28086.png
EDIT: Whoops, it should be f(t) instead of f(x). But I'm lazy...

So pictorially speaking, ddxG(x)\frac{d}{dx}G(x) is the "yellow area" as h gets smaller and smaller. Well, if h is super duper small, the yellow area is effectively the same as the value of f evaluated at x (i.e. "the length of the straight line"), right? This is what FTC (part 1, IIRC) is essentially saying - 'a small change in the area under the curve is the function value evaluated at the "variable point"'.

FTC part 2 is simply what you know about definite integrals.

Do we actually care about which is part 1 and which is part 2? Not really tbh...

---

However, I'm more interested in what you actually want to learn about FTC. All FTC part 1 is saying is you can differentiate an integral with the x at the upper limit, and it's as easy as just plugging x into the integrand itself. Similarly, if the upper limit is x^2, you can still do the trick, but don't forget by chain rule, we need to multiply 2x.
(edited 1 month ago)
Reply 3
Original post by tonyiptony
Another way of thinking about your version of FTC (I think this is the 3blue1brown Essence of Calculus explanation):
Remember, if you fix the value for x, G(x)=axf(t)dtG(x) = \int^x_a f(t) dt is the area under the curve from a to x (trust me on this for now, but I assume you'd agree), and ddxG(x)\frac{d}{dx}G(x) is the change in G(x) given a small change in x (in the picture, the small change is h - this is what d/dx means by definition in words).
rect28086.png
EDIT: Whoops, it should be f(t) instead of f(x). But I'm lazy...
So pictorially speaking, ddxG(x)\frac{d}{dx}G(x) is the "yellow area" as h gets smaller and smaller. Well, if h is super duper small, the yellow area is effectively the same as the value of f evaluated at x (i.e. "the length of the straight line"), right? This is what FTC (part 1, IIRC) is essentially saying - 'a small change in the area under the curve is the function value evaluated at the "variable point"'.
FTC part 2 is simply what you know about definite integrals.
Do we actually care about which is part 1 and which is part 2? Not really tbh...
---
However, I'm more interested in what you actually want to learn about FTC. All FTC part 1 is saying is you can differentiate an integral with the x at the upper limit, and it's as easy as just plugging x into the integrand itself. Similarly, if the upper limit is x^2, you can still do the trick, but don't forget by chain rule, we need to multiply 2x.

Yeah, I understand what the theorem is essentially saying but I am asked to explain it in my own words.

I was confused because I had no idea what the antiderivative was until I looked it up today lol.

I have essentially said discovered by Newton and Leibniz and explored by others, it bridges a connection between differentiation and integration. Part one basically shows that the antiderivative of function f(t) with upper limit x within an interval on a continuous function, is F(x), and when you differentiate F(x)= the original function. So, F(x)=f’(x) and the second part just explain the evaluation of it.

I think that’s essentially what it states
Reply 4
Original post by KingRich
Yeah, I understand what the theorem is essentially saying but I am asked to explain it in my own words.
I was confused because I had no idea what the antiderivative was until I looked it up today lol.
I have essentially said discovered by Newton and Leibniz and explored by others, it bridges a connection between differentiation and integration. Part one basically shows that the antiderivative of function f(t) with upper limit x within an interval on a continuous function, is F(x), and when you differentiate F(x)= the original function. So, F(x)=f’(x) and the second part just explain the evaluation of it.
I think that’s essentially what it states


It would probably help to see a pic of what you wrote, but the bold should be the other way round so
F'(x) = f(x)
Reply 5
Original post by mqb2766
It would probably help to see a pic of what you wrote, but the bold should be the other way round so
F'(x) = f(x)

Yeah, sorry that’s what I meant. I’m not sure if uploading here would risk the chance of it being picked up as plagiarism?

If I screen shot in sections, that could work?
Reply 6
Original post by KingRich
Yeah, sorry that’s what I meant. I’m not sure if uploading here would risk the chance of it being picked up as plagiarism?
If I screen shot in sections, that could work?

Wasnt sure if it was a typo. All the info is in the previous link and as they say, instead of viewing integration geometrically (so the area "under a curve") or as a riemann sum, the ftc says its the inverse of differentiation (and vice versa).

If there is anything youre unsure about, just ask about that bit if you dont want to run the plagiarism risk.
(edited 1 month ago)
Reply 7
Original post by mqb2766
Wasnt sure if it was a typo. All the info is in the previous link and as they say, instead of viewing integration geometrically (so the area "under a curve") or as a riemann sum, the ftc says its the inverse of differentiation (and vice versa).
If there is anything youre unsure about, just ask about that bit if you dont want to run the plagiarism risk.

Honestly, i think it was the technical words that were being used. Now I know what the words mean and after re-reading and re-reading, it’s actually quite straightforward.
Original post by KingRich
Honestly, i think it was the technical words that were being used. Now I know what the words mean and after re-reading and re-reading, it’s actually quite straightforward.

I posted quite a long reply when you previously asked about this (that I'm guessing didn't really help you).

I don't want to revist that in great detail, but to understand why the FTC is important, you need to understand that fundamentally (see what I did there), integration is about measuring the area under a curve. To put it another way, the "true" definition of abf(x)dx\displaystyle \int_a^b f(x) \, dx is that it's the area under the curve y=f(x) between x=a and x=b.

So finding an integral from first principles means making a successive set of approximations that converge to the true area. Conceptually, it's like using squared paper and counting the squares under the curve, and then using smaller and smaller squares to get a better result. It's quite tricky to do for even relatively common functions like x^n.

It's the FTC that says "if you can find a function g whose derivative is f, then abf(x)dx=g(b)g(a)\displaystyle \int_a^b f(x)\,dx = g(b) - g(a)" (*)

As a consequence of the FTC, if you don't care about your students knowing how to do it "from first principles" (and in general, you don't care), you teach them to integrate using (*). Which is perfectly practical, but it means the FTC seems "trivial" (because you're using a definition of integral that makes it trivial).

To look at it another way, if you define integrals using (*), then it's the FTC that actually tells you that integrating does give you the area under the curve (as opposed to something else).
(edited 1 month ago)

Quick Reply