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I want to understand the connection between the primitive function or antiderivative and the definite integral.

My problem with this is the independent variable called t in the formula for the first part of the Fundamental Theorem of Calculus.

Here's a composite of the answers I've already seen for this question. Because of t I don't understand it:

The primitive is a function $F(x)$ such that

$F′(x) = f(x)$

The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this:

$$F(x) = \int_a^x f(t) dt $$ or “$F(x)$ is a primitive function of $f(x)$. The lower bounds a is fixed. The upper bounds $x$ is variable.”

You can write $\int f(x) dx$ as $$\int_a^x f(t) dt + C$$

A complete analysis of this problem is given in Richard Courant’s calculus book (linked below) page 109+.

Some of the sites I’ve seen: http://ia700700.us.archive.org/34/items/DifferentialIntegralCalculusVolI/Courant-DifferentialIntegralCalculusVolI.pdf

http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.html

web.utk.edu/~wneilson/mathbook.pdf page 153-154

www.physicsforums.com/showthread.php?t=212449&highlight=x+dillema

www.jirka.org/diffyqs/htmlver/diffyqsse3.html

math.stackexchange.com/questions/105937/what-does-integration-do

www.math.hmc.edu/calculus/tutorials/fundamental_thm/

www.intuitive-calculus.com/fundamental-theorem-of-calculus.html

  • 6
    Too long; didn't read. What's the question?2012-02-14
  • 3
    @GerryMyerson, +1, If you want quicker answers (& possibly, better answers). Summarize what you wish to ask.2012-02-14
  • 4
    I also don't understand the question. The answer to the question asked in the title is simply "the fundamental theorem of calculus." Is that what your question is about?2012-02-14
  • 0
    Let's discretize the situation and perhaps this will help you tell us what your real confusion is. One can replace functions on the real line by sequences $a_n, n \in \mathbb{Z}$. The analogue of differentiation is taking finite differences $b_n = a_{n+1} - a_n$. The analogue of definite integration is taking sums $\sum_{i=n}^m a_i$, and the analogue of indefinite integration is taking indefinite sums $b_n = \sum_{i=n_0}^n a_i$. The "fundamental theorem of finite differences" says the obvious thing and is straightforward to prove. What is confusing about this situation?2012-02-14
  • 1
    Have you seen [this answer](http://math.stackexchange.com/a/15302/742)?2012-02-14

3 Answers 3