Lesson 20.pdf - 20 The Fundamental Theorem of Calculus August 5 2019 The fundamental theorem of Calculus —– Version 1 The “integral” and

Lesson 20.pdf - 20 The Fundamental Theorem of Calculus...

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20. The Fundamental Theorem of Calculus August 5, 2019 The fundamental theorem of Calculus —– Version 1 The “integral” and “differen- tial” form a pair of “inverse operators” to each other. Namely, given a function, if one takes integral first, and derivative second, then one gets back the function: d dx Z x a f ( t ) dt = f ( x ) . Conversely, if one takes integral first, and derivative second, then one essentially also gets back the function (plus a constant f ( a )): Z x a f 0 ( t ) dt = f ( x ) - f ( a ) . More precisely, we have Theorem 0.1 and 0.4, called the fundamental theorem of Calculus . Theorem 0.1 (Fundamental theorem of Calculus —– Version 1) Let f be integrable on [ a, b ] . Define F ( x ) := Z x a f ( t ) dt. Then 1. F is uniformly continuous on [ a, b ] . 2. If f is continuous at c [ a, b ] , then F 0 ( c ) = f ( c ) . 172
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Proof: 1. We show that F is uniformly continuous on [ a, b ]. Namely, to show that for any > 0, there exists δ > 0 such that | F ( x 1 ) - F ( x 2 ) | < , ∀| x 1 - x 2 | < δ. (1) We consider for any x 1 , x 2 [ a, b ] and assume x 1 < x 2 , | F ( x 1 ) - F ( x 2 ) | = Z x 1 a f ( t ) dt - Z x 2 a f ( t ) dt = Z x 2 x 1 f ( t ) dt Z x 2 x 1 | f ( t ) | dt. Here we used Theorem 0.8 and Corollary 0.11 in § 19. Since f is continuous on [ a, b ] and hence bounded, we have | f ( x ) | ≤ M for any x [ a, b ]. Then we consider | F ( x 1 ) - F ( x 2 ) | ≤ Z x 2 x 1 | f ( t ) | dt M | x 2 - x 1 | . From M | x 1 - x 2 | < , we solve it to take 0 < δ < M , so that (1) holds. 2. To show that for any > 0, there exists δ > 0 such that F ( x ) - F ( c ) x - c - f ( c ) < , 0 < | x - c | < δ. Consider the following (assume x > c ) F ( x ) - F ( c ) x - c - f ( c ) = F ( x ) - F ( c ) x - c - f ( c )( x - c ) x - c = 1 | x - c | Z x a f ( t ) dt - Z c a f ( t ) dt - f ( c ) Z x c dt = 1 | x - c | Z x c f ( t ) dt - Z x c f ( c ) dt = 1 | x - c | Z x c [ f ( t ) - f ( c )] dt 1 | x - c | Z x c | f ( t ) - f ( c ) | dt. Here we used the formula R b a f = R c a f + R b c f for any three real numbers a, c, b (in any order). Since f is uniformly continuous, for the > 0, there exists δ > 0 such that | f ( x 1 ) - f ( x 2 ) | < for any x 1 , x 2 [ a, b ] with | x 1 - x 2 | < δ . Then if | x - c | < δ , we should have F ( x ) - F ( c ) x - c - f ( c ) R x c | f ( t ) - f ( c ) | dt | x - c | < | x - c | | x - c | = . The desired inequality holds. Application in differential equations One application for the above theorem is to solve some differential equations.
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