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Proof of fundamental theorem of calculus

AP.CALC:
FUN‑6 (EU)
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FUN‑6.B (LO)
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FUN‑6.B.1 (EK)
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FUN‑6.B.2 (EK)
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FUN‑6.B.3 (EK)
The fundamental theorem of calculus is very important in calculus (you might even say it's fundamental!). It connects derivatives and integrals in two, equivalent, ways:
I.ddxaxf(t)dt=f(x)II.ab ⁣ ⁣f(x)dx=F(b) ⁣ ⁣ ⁣F(a)\begin{aligned} I.&\,\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x) \\\\ II.&\,\displaystyle\int_a^b\!\! f(x)dx=F(b)\!-\!\!F(a) \end{aligned}
The first part says that if you define a function as the definite integral of another function f, then the new function is an antiderivative of f.
The second part says that in order to find the definite integral of f between a and b, find an antiderivative of f, call it F, and calculate F, left parenthesis, b, right parenthesis, minus, F, left parenthesis, a, right parenthesis.
The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

First, we prove the first part of the theorem.

Khan Academy video wrapper
Proof of fundamental theorem of calculusSee video transcript

Next, we offer some intuition into the correctness of the second part.

Khan Academy video wrapper
Intuition for second part of fundamental theorem of calculusSee video transcript

Finally, we prove the second part of the theorem based on the first part.

Khan Academy video wrapper
The fundamental theorem of calculus and definite integralsSee video transcript

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  • piceratops ultimate style avatar for user Burhanuddin Salim
    In the mean value theorem for integrals proof Sal uses the fundamental theorem of calculus and here in the first part he uses the mean value theorem. Isn't that a circular argument because it says that MVT is true from FTC and FTC is true from MVT?
    (11 votes)
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  • old spice man green style avatar for user Chico Vampiro
    the first proof... thought a little about it and came up w/ an possible other way to solve it:

    take it from the Riemann definition (right sum approx) by
    b=x (upper bound)
    d/dx(int[a;b]f(x)dx = d/dx(f(a+dx)dx+f(a+2dx)dx...n times...+f(b)dx)
    as f(a+dx)dx+f(a+2dx)dx... etc are not x-dependent, the derivatives of all go to 0, so
    d/dx(int[a;b]f(x)dx=d/dx(f(b)dx)=f(b)
    seems reasonable for me, as it all relies on x being both, the upper bound (b) and the variable, as if i just put x=b, considering b as "the max{x}" in the defined interval [a;b]; besides it avoids all the f(t)dt mess that seems confusing.

    so, i want to discuss this, even (probably), why is it wrong (didn't saw it anywhere), but seems quite intuitive for me, for dF/dx is just the last d(right area)=f(b)dx/dx + the local linearity stuff.
    (13 votes)
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    • male robot hal style avatar for user Kamal Mohamed
      Looks like you did a good job there, but I still can't understand what you did because it's hard to understand it when it's written in that way. Can you, or anyone else, please write it down on a piece of paper and upload a photo of it on the internet so everybody can comprehend what's going on?
      Thank you for your addition.
      (2 votes)
  • leaf green style avatar for user filmmoviefilm001
    I understood the first part of the theorem this way. Integrating from a to x(between a and b) is like moving x-pivot from a to b. Along the way x -pivot moves, we add the y- value that we get from the function f(t) (Since F(t) is an integration.) So the F(t)'s derivative of the moment we enter the arbitrary x into f(t) must be +f(x), because F(x) is adding f(x) at that moment, and the derivative describes the rate of change at that moment. Is my intuition correct?
    (7 votes)
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  • spunky sam blue style avatar for user Avi Perl
    What about indefinite integrals? Those don't correspond to any area of a curve, yet they equal the anti-derivative of the function. How does that work?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      The indefinite integral is basically to have a general antidriative. Since the antiderivative is the integral. The problem is the derivative of f(x) is the same as the derivative of f(x)+1 or plus or minus any constant. So the general form as I called it was given a C to represent you could take the derivative of this where C is any real number, and you would get the original function.

      Let me know if that didn't help.
      (2 votes)
  • duskpin ultimate style avatar for user Maria Șandru
    What is that dt and why does it matter?
    (5 votes)
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  • male robot hal style avatar for user Marcus Dentrey
    I'm confused why we needed to prove that d/dx of F is f, when F was defined during creation as the area under f. When we (or Leibnitz I suppose I should say) specified the "width" as dx, doesn't that in itself say that the derivative of F, (the rate at which the area, F, is changing at x) is given by f(x)? Stated another way, by specifying the width as dx, which is negligible in the area calculation by design, doesn't our area change by f(x)? Perhaps this is not proof because it's recursive?
    (2 votes)
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  • leaf yellow style avatar for user Hayoun Lee
    Would there be any critical errors in this analogy? (I came up with this to understand those theorems more intuitively.)

    f = F'
    f = productivity (speed in producing value)
    F = wealth (accumulated value)
    C = Inherited wealth
    If productivity is static, wealth increases linearly
    if productivity increases, wealth increases exponentially
    -> INVEST IN EDUCATION!
    (3 votes)
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  • starky ultimate style avatar for user Cameron Jenkins
    This would be a lot easier if different letters were used for the function. Sometimes it's hard to tell if he means f or F.
    (3 votes)
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  • duskpin ultimate style avatar for user Tong Qiu
    Since the MVT requires the second FTC ( based on the video https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-integration-new/ab-8-1/v/mean-value-theorem-integrals ), and the first FTC requires the MVT, shouldn't we prove the second FTC before proving the first one?

    Also, how exactly is the second FTC based on the first one? I can't see how the first one would be necessary to prove the second one, and if it were to be necessary we would be caught in a circular reasoning.
    (2 votes)
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    • leaf green style avatar for user kubleeka
      In the linked video, Sal is pointing out a connection between the MVT and integration. He is not proving the MVT.

      To actually prove the MVT doesn't require either fundamental theorem of calculus, only the extreme value theorem, plus the fact that the derivative of a function is 0 at its extrema (when the derivative exists). That should defuse any fears of circular reasoning.
      (1 vote)
  • female robot grace style avatar for user SULAGNA NANDI
    Where in the AP calc BC playlist is the video for the MVT of definite integrals? I've gone through all the material before this and I haven't seen it yet, but Sal talks about it in the first video here.
    (1 vote)
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