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File: Chain Rule Pdf 170099 | Lecture11
outline chain rule dierentiation rules s sial dept of mathematics lums fall 2020 2021 outline derivatives of other trig functions derivatives of trig functions s sial dept of mathematics lums ...

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                                         Math 141: Lecture 11
            The Fundamental Theorem of Calculus and integration methods
                                                      Bob Hough
                                                 October 12, 2016
                Bob Hough                            Math 141: Lecture 11                        October 12, 2016     1 / 36
   First Fundamental Theorem of Calculus
    Theorem (First Fundamental Theorem of Calculus)
    Let f be a function that is integrable on [a,b]. Let c be such that
    a ≤ c ≤ b and define, for a ≤ x ≤ b,
                                                A(x) = Z x f(t)dt.
                                                               c
                              ′
    The derivative A (x) exists at each point x ∈ (a,b) where f is continuous
    and A′(x) = f(x).
                Bob Hough                            Math 141: Lecture 11                        October 12, 2016     2 / 36
   First Fundamental Theorem of Calculus
    Proof.
            Let f be continuous at x. Given ǫ > 0 choose δ > 0 such that if
            |h| < δ, |f (x + h) − f (x)| < ǫ.
            Write, for h 6= 0,
                  A(x +h)−A(x)=Z x+hf(t)dt −Z xf(t)dt = Z x+hf(t)dt.
                                                   c                        c                    x
            Thus
            A(x +h)−A(x) = 1 Z x+hf(t)dt = f(x)+ 1 Z x+h(f(t)−f(x))dt.
                        h                   h x                                     h x
            For |h| < δ the last integral is bounded in size by ǫ, from which the
            limit follows.
                Bob Hough                            Math 141: Lecture 11                        October 12, 2016     3 / 36
   Zero-derivative Theorem
    Theorem
    If f ′(x) = 0 for each x in an open interval I, then f is constant on I.
    Proof.
    Let x 6= y in I. Since f is differentiable, by the Mean Value Theorem there
    is z between x and y such that 0 = f ′(z) = f(y)−f(x).
                                                                               y−x
                Bob Hough                            Math 141: Lecture 11                        October 12, 2016     4 / 36
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