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The Fundamental Theorem of Calculus [1] Guangliang Zhao 1 Review (i) Definite Integral: Limit of Riemann Sum. Three steps: Divide-Sum-Limit (DSL). §5.2, p372. ∫x 2 2 2 Ex: 0 tdt = x2 − 02 = x2 . (ii) Derivative: f ′ (a) is the instantaneous rate of change of y = f (x) with respect to x when x = a. (TANGENT). §2.7, p148. Ex: ( x2 )′ = x. 2 (iii) Antiderivative F of f is defined by: F ′ = f . Q: Definite integral arose from the area problem, and Derivative arose from the tangent problem. What’s the deep relationship between definite integral and derivative? 2 2.1 The Fundamental Theorem of Calculus (FTC) Part 1 Suppose f is a continuous function on [a, b], x varies between a and b. FTC deals with the function ∫ x f (t)dt, g(x) = a which can be interpreted as the area under the graph of f from a to x, and the blue part in Figure 1. Here in the graph f happens to be a positive function. To compute g ′ (x), let’s use the definition of a derivative. For h > 0, g(x + h) − g(x) is obtained by subtracting areas, so it is the area under the graph of f from x to x + h (the red area in Figure 1). For small h, the area is approximately equal to the area of the rectangle with height f (x) and width h: g(x + h) − g(x) ≈ hf (x) so g(x + h) − g(x) ≈ f (x). h Therefore we expect that g ′ (x) = lim h→0 g(x + h) − g(x) = f (x). h 1 Figure 1: Theorem 2.1 (The Fundamental Theorem of Calculus, Part 1.) If f is continuous on [a, b], then the function g defined by ∫ x g(x) = f (t)dt a ≤ x ≤ b a is continuous on [a, b] and differentiable on (a, b), and g ′ (x) = f (x). Proof. If x and x + h are in (a, b), then ∫ ∫ x f (t)dt − f (t)dt ) ∫ x (a∫ x ∫ ax+h f (t)dt f (t)dt − = f (t)dt + a x a ∫ x+h f (t)dt = g(x + h) − g(x) = x+h (by Property 5 of integrals) x so for h ̸= 0, g(x + h) − g(x) 1 = h h ∫ x+h f (t)dt. x Now let’s assume that h > 0. Since f is continuous on [x, x+h], by Extreme Value Theorem, there are numbers u and v in [x, x + h] such that f (u) = m and f (v) = M , where m and M are the absolute minimun and maximum values of f on [x, x+h]. By Property 8 of integrals, we have ∫ x+h mh ≤ f (t)dt ≤ M h x that is ∫ f (u)h ≤ x+h f (t)dt ≤ f (v)h x Since h > 0, we can divide this inequality by h: 1 f (u) ≤ h ∫ x+h x 2 f (t)dt ≤ f (v) replace the middle part by g, we get f (u) ≤ g(x + h) − g(x) ≤ f (v) h (2.1) Letting h → 0. Then u → x and v → x, since u and v lie between x and x + h. Therefore lim f (u) = lim f (u) = f (x) u→x h→0 and lim f (v) = lim f (v) = f (x) h→0 v→x because f is continuous at x. By Squeeze Theorem and (2.1) we conclude that g ′ (x) = lim h→0 g(x + h) − g(x) = f (x). h (2.2) If x = a or b, then (2.2) can be interpreted as a one-sided limit. Then Theorem 2.8.4 shows that g is continuous on [a, b]. Remark 2.2 This theorem first tells us that definite integral of f is one of it’s infinitely many antiderivatives. That is because g is a definite integral by definition, and the conclusion g ′ = f indicates that g is an antiderivative of f . Remark 2.3 Using Leibniz notion for derivatives, we can write FTC1 as ∫ x d f (t)dt = f (x) dx a (2.3) when f is continuous. Roughly speaking, 2.3 says that if we first integrate f and then differentiate the result, we get back to the original function f . This shows that an antiderivative can be reversed by a differentiation, and it also guarantees the existence, continuity, differentiability of antiderivatives ∫x d ) of the area ( a ), for continuous functions. In other words, the instantaneous rate of change ( dx which located under the curve f from a to x as shown in Figure 1, evaluated at x, is exactly f evaluated at point x. ∫x√ Example 2.4 Find the derivative of the function g(x) = 0 1 + t2 dt. Solution. Since f (t) = Example 2.5 Find d dx √ 1 + t2 is continuous, by FTC part 1, √ g ′ (x) = 1 + x2 . ∫ x4 1 sec tdt. Solution. Let u = x4 . Then ∫ u ∫ x4 d d sec tdt dx 1 sec tdt = dx 1 [∫ ] u du d sec tdt (by the Chain Rule) = du 1 dx du (by FTC1) = sec u dx 4 = sec(x ) · 4x3 References [1] James SteWart, Calculus, Early Transcendentals, 7th ed., Brooks/Cole. 3