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2nd fundamental theorem of calculus

by snakehunter
Tags: calculus, fundamental, theorem
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snakehunter
#1
Apr26-04, 03:40 PM
P: 2
Can some on pleases explain this too me. I have an AP book, and i am to do a few problems out of it for class, and but cant find it in there ANY WHERE. Any help would be superb!
-Jacob
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pig
#2
Apr26-04, 04:07 PM
P: 87
Mathworld says:

http://mathworld.wolfram.com/SecondF...fCalculus.html

In my book, this is simply called "the fundamental theorem of calculus", and mathworld's first t. of c. is mentioned later than this, so I can understand your confusion.
Ebolamonk3y
#3
Apr26-04, 04:32 PM
P: 204
Finney's book has this backwards... The second F of C is the Integral Evaluation Theorem... Which is the mathworld's 1st...

Its hand stuff!

Math Is Hard
#4
Apr26-04, 04:55 PM
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2nd fundamental theorem of calculus

I've seen it both ways in several books. Luckily, no teacher has ever asked me to state FTC #1 or state FTC #2, they've only asked me to be able to use them.
Did you try any of the problems yet?
JonF
#5
Apr26-04, 05:29 PM
P: 617
Do you need to understand the proofs? Or just understand the idea?

FTC#2 says
[itex] \int^b_a f(x)dx = F(b) - F(a) [/itex]

where the anti-derivative of f(x) is F(x)

So what you are probably using this Thrm for is evaluating definite integrals (ones with A and B stated).

So what you do is find the anti-derivative of f(x) i.e. F(x), and then evaluate that anti-derivative at a and b, then take the absolute value of their difference.

For example if we wanted to evaluate:

[itex] \int^5_1 3x^2dx [/itex]

we would first find the anti derivative of 3x^2.

Which is x^3.

Then we would evaluate x^3 at 1 and 5 which gives us, 1 and 125. We subtract 1 from 125 and get 124, hence:

[itex] \int^5_1 3x^2dx = 124[/itex]
snakehunter
#6
Apr26-04, 06:15 PM
P: 2
thanx you guys. I know how to do those with my eyes closed, its just that my AP book has a real habit of doing things w/o thouroughly explaining them. thanx again ill tell my friends about this site!
-Jacob
mathwonk
#7
Sep21-04, 09:07 PM
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2nd FTC: If f is any riemann integrable function on the closed bounded interval [a,b], and G is a Lipschitz - continuous function such that for every point x where f is continuous, G is diiferentiable at x with G'(x) = f(x), then the integral of f from a to b, equals G(b)-G(a).

Recall that G is lipschitz continuous on [a,b] if there exists a constant K such that for all points u,v in [a,b] we have |G(v) - G(u)| <= K|v-u|.
JasonRox
#8
Sep22-04, 09:21 PM
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Is the first theorem just:

y=x^n
dy/dx=nx^n-1

Is that it?

I wrote general proof for the derivative of a function x^n. Using limits of course, and I ended with nx^n-1, which is what you are suppose to get.

I never seen the proof yet, and it would be great to do it independently.

So, what is the 1st Fundamental Theorem(in most books)?
mathwonk
#9
Sep22-04, 09:27 PM
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in most books it says that if f is continuous on [a,b] and G(x) is the integral of f from a to x, then G is differentiable on [a,b] and G'(x) = f(x) for every x in [a,b].

The more general statement is that if f is a Riemann integrable function on [a,b] and G(x) again is the injtegral of f from a to x, then G is Lipschitz continuous, and G is differentiable with G'(x) = f(x) at those points where f is continuous.

Then to derive the 2nd thm from the first you need the generalized mean value theorem, that a function G which is lipschitz continuous and has derivative equal to zero almost everywhere (i.e. except on a set of measure zero) is constant.

in most books the 2nd thm just says that if f is continuous on [a,b] and G is continuous on [a,b] with G'(x) = f(x) for all x in (a,b), then the integral of f from a to b, equals G(b)-G(a).
JasonRox
#10
Sep23-04, 02:14 PM
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The Riemann Sum?

You can find the value under a curse with it.


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