Question

can anyone please provide me the proof of fundamental theorem of integral calculus??

Answer 1

The fundamental theorem of LINE integrals?

Answer 2

or FToC Part 2?

Answer 3

fundamental theorem of integral calculus using primitive of the function

Answer 4

I think you're referring to FToC part II.
Ok let me start off. I will see what i can remember if i remember it correctly. I will do it by parts

Answer 5

Yes, @prasadchoklet is referring to second part, where you put the integral of \(f\) in terms of the endpoints of \(F\).

Answer 6

I think by definition its said that if you have a continuous function, for instance, f(x) on a given interval, [a,b], then F(x) is an antiderivative of your function f(x). Explained by:
\[\int\limits_{a}^{b}f(x)dx = F(b)-F(a)\]
meaning that its evaluated on that interval

Answer 7

or maybe it is
\[\frac{d}{dx}\int_a^xf(t)dt=f(x)\] not sure which is the first or the second
in any case it is in any calc 1 text

Answer 8

First
\[g(x) = \int\limits_{a}^{x}f(t)dt\]
from part 1, you know g'(g)=f(x) so this means it's an antiderivative.
This means
g'(x)=F'(x)
By the mean value theorem, g(x) and F(x) cannot differ by more than an additive constant on (a,b). That means a<x<b, which gives you:
\[F(x)=g(x)+K\] where K is some arbitrary constant.
Since g(x) and F(x) are both continuous on the given interval [a,b, then that means if you take the lim as x->a from the right and x->b from the left, you can see that it holds true if x=a and x=b

Answer 9

@satellite73 No, I just checked online. What you're referencing is the first part. Doesn't matter anyway though. Not like order matters.

Answer 10

So, for a<x<b you know that F(x)=g(x)+K
hence, \[F(b)-F(a)=(g(b)+K)-(g(a)+K) = g(b)-g(a)\]

Answer 11

\[\int\limits_{a}^{b}f(t)dt + \int\limits_{a}^{a}f(t)dt = \int\limits_{a}^{b}f(t)dt+0 = \int\limits_{a}^{b}f(x)dx\]

Answer 12

iterchanging variables doesnt matter as you might recognize that when u do a couple integrals. You can have theta, or x, or r, etc.,

Answer 13

I was using the mean value theorem to prove that If f'(x)=g'(x) for all x in an interval (a,b), then in the interval, f(x)=g(x)+K
where K is some constant.