Chapter Integration Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 7.3, Slide Section 7.3 The Fundamental Theorem of Calculus Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 7.3, Slide Theorem 7.3.1 The Fundamental Theorem of Calculus I (Part 1) F ( x) = Let f be integrable on [a, b] For each x ∈ [a, b], let x ∫a f (t ) dt Then F is uniformly continuous on [a, b] Proof: Since f is integrable on [a, b], it is bounded there That is, there exists B > such that | f (x) | ≤ B for all x ∈ [a, b] To see that F is uniformly continuous on [a, b], suppose ε > If x, y ∈ [a, b] with x < y and | x – y | < ε /B, then y F ( y ) − F ( x) = ∫a ≤ y Corollary 7.2.8 ∫x Theorem 7.2.6 x f (t ) dt − ∫ f (t ) dt = a f (t ) dt ≤ y ∫x B dt y ∫x f (t ) dt = B ( y − x) < ε Thus F is uniformly continuous on [a, b] Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 7.3, Slide The Fundamental Theorem of Calculus I (Part 2) Let F ( x) = x ∫a f (t ) dt If f is continuous at c ∈ [a, b], then F(x) is differentiable at c and F ′(c) = f (c) Given any ε > there exists a δ > such that | f (t) – f (c)| < ε whenever Proof: Now for all x ≠ c we have t ∈ [a, b] and | t – c | < δ f (c ) = x f (c) dt ∫ c x−c Thus for any x ∈ [a, b] with < | x – c | < δ we have c F ( x ) − F (c )  x  − f (c ) − f (c) = f ( t ) d t − f ( t ) d t ∫a  x−c x − c  ∫a = x x f ( t ) dt − f (c) dt ∫ ∫ c c x−c x−c ≤ x−c Copyright © 2013, 2005, 2001 Pearson Education, Inc x ∫c | f (t ) − f (c) | dt < = x−c x ∫c [ f (t ) − f (c)] dt ε | x − c | = ε x−c Section 7.3, Slide F ( x ) − F (c ) − f (c ) < ε Thus we have x−c Since ε F ( x ) − F (c ) = F ′(c ) ♦ x→c x−c f (c) = lim > was arbitrary, we must have d dx In the Leibnitz notation: (∫ ) x f (t )dt = f ( x ) a Example 7.3.2 If F ( x) = x ∫0 + t dt or d dx Copyright © 2013, 2005, 2001 Pearson Education, Inc for x ≥ 0, then (∫ x ) + t dt = F ′( x) = + x3 + x3 Section 7.3, Slide Corollary 7.3.3 Let f be continuous on [a, b] and let g be differentiable on [c, d], where g([c, d]) ⊆ [a, b] Define F ( x) = g ( x) ∫a f (t ) dt for all x ∈ [c, d] F ′( x) = [( f og )( x)]g ′( x) Then F is differentiable on [c, d] and (The proof is a straightforward application of the chain rule.) Example 7.3.4 If F ( x) = x2 ∫0 + t dt for x ≥ 0, F ′( x) = ( Copyright © 2013, 2005, 2001 Pearson Education, Inc then ) + ( x )3 (2 x) = x + x Section 7.3, Slide The Fundamental Theorem of Calculus II Theorem 7.3.5 b ∫a f ′ If f is differentiable on [a, b] and f ' is integrable on [a, b], then Proof: Let P = {x0, x1, …, xn} be any partition of [a, b] By applying the mean value = f (b) − f (a ) theorem to each subinterval [xi – 1, xi], we obtain points ti ∈ (xi – 1, xi) such that f ( xi ) − f ( xi −1 ) = Thus we have [ f '(ti )] ( xi − xi −1) n ∑ [ f ( xi ) − f ( xi −1)] f (b) − f (a ) = = i =1 n ∑ [ f '(ti )] ( xi − xi −1) i =1 All but the first and last terms cancel Since mi( f ') ≤ f '(ti) ≤ Mi( f ') for all i, it follows that L( f ', P) ≤ f (b) – f (a) ≤ U( f ', P) This holds for each partition P, so we also have But f ' L( f ' ) ≤ f (b) – f (a) ≤ U( f ' ) L( f ′) = U ( f ′) = is assumed to be integrable on [a, b], so Thus Copyright © 2013, 2005, 2001 Pearson Education, Inc f (b) − f (a ) = b ∫a f ′ b ∫a f ′ ♦ Section 7.3, Slide Definition 7.3.8 lim c → a + ∫ cb f Let f be defined on (a, b] and integrable on [c, b] for every c ∈ (a, b] If exists, then the improper integral of f on (a,  b], denoted by ∫ ab f , is given by ∫ ab f = limc → a + ∫ cb f If L = lim c →is a +finite f then the improper integral ∫ cb number, is said to converge to L ∫ ab f If L = ∞ (or – ∞), then the integral is said to diverge to ∞ (or – ∞) Example 7.3.9(a) Let f (x) = x – 1/3 for x ∈ (0,  1] Then f is not integrable on [0, 1], since it is not bounded there But for each c ∈ (0, 1] we have ∫c1 x −1/3dx = = 3/2 (1 − c ) ∫ 01 x −1/3dx = lim c → 0+ 32 (1 − c3/2 ) = 32 Taking the limit as c → +, we have So the improper integral 2/3  x  c ∫ x −1/3 dx converges0to Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 7.3, Slide Example 7.3.9(b) Let ∫ c 1x dx = ln1 − ln c = − ln c g (x) = 1/x for x ∈ (0, 1] Then for each c ∈ (0, 1] we have Since lim c → + (– ln c) = ∞, the improper integral diverges to dx we write ∫ 10∞1x and ∫01 1x dx = ∞ We can also define an improper integral over an unbounded interval Definition 7.3.10 c f be defined on [a, and integrable on [a, for every c > a If ∞ ∫a f exists, then the improper integral of f ∞ c f = lim ∫ c →∞ a f ∫a Let  ∞)  c] on [a, ∞), denoted by Copyright © 2013, 2005, 2001 Pearson Education, Inc limc →∞ ∫ a f , is given by Section 7.3, Slide ... differentiable on [a, b] and f ' is integrable on [a, b], then Proof: Let P = {x0, x1, …, xn} be any partition of [a, b] By applying the mean value = f (b) − f (a ) theorem to each subinterval [xi... improper integral diverges to dx we write ∫ 10∞1x and ∫01 1x dx = ∞ We can also define an improper integral over an unbounded interval Definition 7.3.10 c f be defined on [a, and integrable on [a,... +finite f then the improper integral ∫ cb number, is said to converge to L ∫ ab f If L = ∞ (or – ∞), then the integral is said to diverge to ∞ (or – ∞) Example 7.3.9(a) Let f (x) = x – 1/3 for x