Calc_Chap08

Calc_Chap08 - MATH1131 – Mathematics 1A Calculus Chapter...

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Unformatted text preview: MATH1131 – Mathematics 1A Calculus Chapter 08 Integration Dr. Thanh Tran School of Mathematics and Statistics The University of New South Wales Sydney, Australia 1 2 3 4 5 6 7 8 Riemann sums and Riemann integrals Riemann integrals and areas Basic properties Fundamental Theorems of Calculus Indefinite integrals Integration by substitution Integration by Parts Improper Integrals 1 2 Basic idea Approximate area under the graph of a (continuous) function y = f (x ) by sum of approximating rectangles. Integration is concerned with the geometric problem of finding the area of a given region. We start by trying to calculate the area between the x axis and the curve y = f (x ) for a ≤ x ≤ b , where f (x ) < M , i.e. f is bounded. (In this course we will always assume this!) a and b are finite. (Improper integrals will be introduced to remove this restriction!) a = x0 x1 x2 ... xi −1 xi . . . xn−1 xn = b 3 Riemann sums and Riemann integrals 4 Definition 2 (Riemann sum) The sum Definition 1 (Partition) A partition of an interval [a, b ] is a set P = {x0 , x1 , x2 , . . . , xn } of points sucht that a = x0 < x1 < x2 < · · · < xn = b . A partition is said to be uniform if xi − xi −1 = In this case xi = a + b−a , n i = 1, . . . , n . ∗ ∗ ∗ SP = (x1 − x0 )f (x1 ) + (x2 − x1 )f (x2 ) + · · · + (xn − xn−1 )f (xn ) n = i =1 (xi − xi −1 )f (xi∗ ) is called a Riemann sum (for f with respect to P ) provided that xi∗ ∈ [xi −1 , xi ] for each i . Definition 3 (Lower and upper sums) i (b − a) , n i = 0, . . . , n . There are two special Riemann sums: If xi∗ is chosen so that f (xi∗ ) is a minimum on [xi −1 , xi ] then we get the lower sum, denoted by S P . If xi∗ is chosen so that f (xi∗ ) is a maximum on [xi −1 , xi ] then we get the upper sum, denoted by S P . Riemann sums and Riemann integrals 5 Riemann sums and Riemann integrals 6 A Riemann sum A lower Riemann sum a = x0 ∗ x1 x1 ∗ x2 x2 ... xi −1 xi . . . xi∗ xn−1 xn = b ∗ xn a = x0 x1 x2 ... xi −1 xi . . . xn−1 xn = b Riemann sums and Riemann integrals 7 Riemann sums and Riemann integrals 8 An upper Riemann sum Definition 4 (Integrable function) We say that f is Riemann integrable on [a, b ] if for every ε > 0 there is a partition P such that 0 ≤ S P − S P < ε. Integral of f over [a, b ] If f is Riemann integrable on [a, b ], there is a unique real number I such that SP ≤ I ≤ SP for all partition P . We then define b b a = x0 x1 x2 ... xi −1 xi . . . xn−1 xn = b I= a f= a f (x ) dx . The function f is called the integrand, and the points a and b are called the limits of the definite integral. Riemann sums and Riemann integrals 9 Riemann sums and Riemann integrals 10 Example 5 Use the formula N S Pn = k2 = k =1 1 N (N + 1)(2N + 1) 6 1 0 = = to show that 1 x 2 dx = . 3 and S Pn = = = Consider the partition P = {x0 , . . . , xn } where xi = i , i = 0, . . . , n . n This is a uniform partition and we will denote it by Pn . Then Riemann sums and Riemann integrals 11 Riemann sums and Riemann integrals 12 Since S Pn − S Pn = 1 if n > we have a partition Pn such that ǫ S Pn − S Pn < ǫ. Hence f (x ) = x 2 is integrable on [0, 1]. (n − 1)(2n − 1) ≤ 6n2 1 0 Example 6 Use the formula N k3 = k =1 12 N (N + 1)2 4 1 . 4 to show that 0 1 Therefore, we have n = 1, 2, 3, . . . . x 3 dx = (n + 1)(2n + 1) x 2 dx ≤ , 6n2 Taking finer and finer partitions we obtain (n − 1)(2n − 1) ≤ n→∞ 6n2 lim or 1 0 x 2 dx ≤ lim (n + 1)(2n + 1) , n→∞ 6n2 Hint: Use the same uniform partition as in the previous example, and check that (n − 1)2 (n + 1)2 S Pn = and S Pn = . 4n 2 4n 2 Then note that 1 (n + 1)2 (n − 1)2 = lim =. 2 n→∞ n→∞ 4 4n 4n 2 lim 13 Riemann sums and Riemann integrals 14 Riemann sums and Riemann integrals Example Therefore, 7 Use first principles to show that x 2 dx = 1 . 3 0 b a 1 Example 8 Let f : [0, 1] → R be defined by f (x ) = 2 1 if x ∈ Q, if x ∈ Q. / C dx = C (b − a). Show that f is not Riemann integrable. Let P be an arbitrary partition of [0, 1]. Then for any i = 1, . . . , n xi −1 ≤x ≤xi min f (x ) = 1 and xi −1 ≤x ≤xi max f (x ) = 2. Hence SP = Let ǫ = 1/2. Then for any partition P there holds S P − S P > ǫ. Hence f is not integrable. Riemann sums and Riemann integrals 15 Riemann sums and Riemann integrals 16 SP = Riemann integrals and areas b If f : [a, b ] → R satisfies f (x ) ≥ 0 for all x ∈ [a, b ], then Definition 9 A function f : [a, b ] → R is said to be piecewise continuous if it is continuous at all except perhaps a finite number of points. Theorem 10 If f : [a, b ] → R is bounded and piecewise continuous, then it is integrable on [a, b ]. is the area under the graph of f from a to b . If f : [a, b ] → R changes sign, then b a f (x ) dx a f (x ) dx = area of regions above the x -axis − area of regions below the x -axis. This quantity is called the signed area under the graph of f from a to b . In general, the area of the region formed by the graph of f , the x -axis, the lines x = a and x = b is b area = a |f (x )| dx . 18 Riemann sums and Riemann integrals 17 Riemann integrals and areas Basic properties 1 4 If f is integrable on [a, b ] and k is a constant then b b If k is a constant then b a 5 kf (x ) dx = k a a f (x ) dx . k dx = k (b − a). If f and g are integrable on [a, b ] then b b b 2 The area bounded by y = f (x ), the x -axis, and the lines x = a and x = b is b a (f + g )(x ) dx = a 6 f (x ) dx + a a g (x ) dx . |f (x )| dx (assuming a < b ). If f is integrable on [a, b ] and f (x ) ≥ 0 for all x ∈ [a, b ] then b a 3 If f is integrable on [a, b ] and a < c < b then b c b 7 f (x ) dx ≥ 0. f (x ) dx = a a f (x ) dx + c f (x ) dx . If f and g are integrable on [a, b ] and f (x ) ≤ g (x ) for all x ∈ [a, b ] then b b a f (x ) dx ≤ g (x ) dx . a 20 Basic properties 19 Basic properties Extending the definition 8 If f is integrable on [a, b ] and m ≤ f (x ) ≤ M for all x ∈ [a, b ], then b So far the integral a f (x ) dx has only been defined when a < b . We now extend this definition. Definition 11 Suppose that a < b and that f is integrable on [a, b ]. Then we define a b a b m(b − a) ≤ 9 a f (x ) dx ≤ M (b − a). If |f | is integrable on [a, b ] then b a b b f (x ) dx := − f (x ) dx a and a f (x ) dx := 0. f (x ) dx ≤ a |f (x )| dx . As a consequence, if f is integrable on an interval containing a, b and c then b c b Proof: See printed notes. a f (x ) dx = a f (x ) dx + c f (x ) dx . That means we can relax on the assumption a < c < b in Property 3. Basic properties 21 Basic properties 22 Theorem 12 (The First Fundamental Theorem of Calculus) Suppose that f is a continuous function on [a, b ]. Then the function F : [a, b ] → R defined by x Proof: We first prove that F is differentiable on (a, b ). x , x + h ∈ (a, b ) there holds x +h For F (x + h) − F (x ) = f (t ) dt . x F (x ) = a f (t ) dt Consider h > 0. Since f is continuous on [x , x + h], it attains a maximum and minimum values on this interval, i.e., mh ≤ f (t ) ≤ Mh Hence mh h ≤ x x +h is continuous on [a, b ], differentiable on (a, b ), and F ′ (x ) = f (x ) We write ∀x ∈ (a, b ). ∀t ∈ [x , x + h]. ∀t ∈ [x , x + h], d dx x f (t ) dt ≤ Mh h f (t ) dt = f (x ). a so that mh ≤ Since h → 0+ 23 Fundamental Theorems of Calculus F (x + h) − F (x ) ≤ Mh . h h →0 It follows that F is an anti-derivative of f . All other anti-derivatives are F (x ) + C for any real number C . Fundamental Theorems of Calculus lim mh = lim+ Mh = f (x ), 24 A more general result F (x + h) − F (x ) = f (x ). lim+ h h →0 A similar argument for h < 0 yields lim F (x + h) − F (x ) = f (x ). h If f is integrable then F is continuous, and F is differentiable where f is continuous. Example 13 If f (x ) = x h → 0− Hence, F is differentiable at x ∈ (a, b ) and F ′ (x ) = f (x ). This implies that F is continuous at x . It remains to show that F is continuous at a and b . By noting that |f (t )| ≤ M for all t ∈ [a, b ] we deduce x x 0 2 x < 0, x ≥ 0, evaluate F (x ) = −1 f (t ) dt and F ′ (x ). |F (x ) − F (a)| = a f (t ) dt ≤ a |f (t )| dt ≤ M |x − a| → 0 as x → a+ . It’s easy to see that F (x ) = 0 2x x <0 x ≥ 0. In other words, limx →a+ F (x ) = F (a). So F is continuous at a. Similar argument holds for b . Fundamental Theorems of Calculus 25 Observe that F ′ (x ) = f (x ) for x = 0, and F is not differentiable at x = 0. Fundamental Theorems of Calculus 26 Theorem 14 (The Second Fundamental Theorem of Calculus) Suppose that f is a continuous function on [a, b ]. If F is an antiderivative of f , i.e., F ′ = f , then b a Example 15 4 Evaluate I = 1 dx √. x f (x ) dx = F (b ) − F (a). x Proof: Defining G(x ) = a √ The integrand f (x ) = 1/ x has a primitive √ F (x ) = 2 x , as F ′ (x ) = 2 1 2 f (t ) dt ∀x ∈ [a, b ] x −1/2 = x −1/2 = f (x ). we deduce G(x ) = F (x ) + C for some constant C . so that C = −F (a). To determine C we substitute x by a to obtain 0 = G(a) = F (a) + C , Therefore, G(b ) = F (b ) − F (a), i.e., b a Fundamental Theorems of Calculus As f is continuous on [1, 4], we can use the SFTC to obtain 4 1 f (x ) dx = F (b ) − F (a). 27 Fundamental Theorems of Calculus dx 4 √ = F (x ) 1 x = F (4) − F (1) √ √ = 2 4 − 2 1 = 2. 28 Notes The First Fundamental Theorem of Calculus states that if f is continuous on [a, b ], then differentiation undoes integration, i.e., if one integrates f and differentiates the result, one obtains f again. In formula f (x ) = d dx x Corollary 16 Suppose that f is continuous on [a, b ] and has a continuous derivative f ′ on (a, b ). Then x f (t ) dt a ∀x ∈ (a, b ). f (x ) = f (a) + a f ′ (t ) dt . The converse is not true. If one differentiates a function f and then integrates the result, one will not obtain f , i.e., in general x Proof: Apply The SFTC for f ′ on [a, x ]. f (x ) = a f ′ (t ) dt , but . . . Fundamental Theorems of Calculus 29 Fundamental Theorems of Calculus 30 Example 17 x Find the following derivatives. (Note: 0 exp(t 2 ) dt cannot be expressed in terms of elementary functions.) x d 1 exp(t 2 ) dt dx 0 2 3 d dx d dx 0 x x x2 exp(t 2 ) dt . exp(t 2 ) dt . Fundamental Theorems of Calculus 31 Fundamental Theorems of Calculus 32 Example 18 Find the continuous function f and the constant c satisfying x 1 Example 19 n f (t ) dt = 0 x x 16 x 18 t f (t ) dt + + + C. 8 9 2 Find lim Sn where Sn = n→∞ i =1 n2 n . + i2 Fundamental Theorems of Calculus 33 Fundamental Theorems of Calculus 34 Indefinite integrals Some formulas x r dx = x r +1 + C , where r is rational, r = −1 r +1 When not interested in a given domain when finding anti-derivatives we write f (t ) dt = F (t ) + C The expression f (t ) dt is called an indefinite integral and the constant C is called the constant of integration. sin x dx = − cos x + C cos x dx = sin x + C f ′ (x ) dx = ln |f (x )| + C f (x ) 1 eax dx = eax + C a dx √ = sin−1 x + C = − cos−1 x + C ′ 2 1−x dx = tan−1 x + C 1 + x2 Indefinite integrals 36 Indefinite integrals 35 Theorem 20 (Change of variable formula) Suppose that g is a differentiable function such that is continuous on (a, b ). If f is continuous on an interval I containing g (a) and g (b ) then b Example 21 g′ Evaluate x2 1 + x 3 dx . f [g (x )]g ′ (x ) dx = a v g (b ) f (u ) du . g (a) Proof Let F (v ) = g (a) f (u ) du . Then F ′ = f and (F ◦ g )′ (x ) = F ′ [g (x )]g ′ (x ) = f [g (x )]g ′ (x ) ∀x ∈ [a, b ]. Hence b a f [g (x )]g ′ (x ) dx = (F ◦ g )(b ) − (F ◦ g )(a) = F [g (b )] − F [g (a)] g (b ) = g (a) Integration by substitution f (u ) du . 37 Integration by substitution 38 Example 22 4 Evaluate 1 sin(π x ) √ dx . x √ Integration by Parts Integrating both sides of d (uv ) dv du =u +v dx dx dx gives uv = or u or, for definite integrals, b u dv dx + dx v v du dx dx du dx dx du dx . dx b a b dv dx = uv − dx b a u a dv dx = uv dx b − b v a We also write u dv = uv − Integration by substitution 39 Integration by Parts v du and a u dv = uv − v du . a 40 Example 23 Evaluate x cos x dx . Example 24 Evaluate sin−1 x dx . Integration by Parts 41 Integration by Parts 42 General rule Integrals of functions with special properties Theorem 25 If f (x ) is even on [−a, a] then a a f (t ) dt = 2 If one of the functions in the product is a polynomial, always chose it to be the function u — the one that gets differentiated. Exception to this is if the other function is ln x or similar — in this special case choose u to be the ln x . −a 0 f (t ) dt If f (x ) is odd on [−a, a] then a f (t ) dt = 0 −a If f is periodic with period T then T +β T f (t ) dt = β 0 f (t ) dt for all β ∈ R. Integration by Parts 43 Integration by Parts 44 Definition 26 If b →∞ a ∞ Example 27 b lim f (x ) dx = L for some real number L, then f is said to be integrable over [a, ∞) and the integral a ∞ a f (x ) dx is said to be convergent. We write f (x ) dx = L, Evaluate the following improper integral or show that it is not convergent. ∞ 1 dx . 1 + x2 0 and call the integral an improper integral. If the limit does not exist we say that the improper integral ∞ f (x ) dx diverges. b b a Similarly, we define −∞ Improper Integrals f (x ) dx = lim a→−∞ a f (x ) dx if the limit exists. 45 Improper Integrals 46 Example 28 Evaluate the following improper integral or show that it is not convergent. ∞ e dx . x 0 Definition 29 A function f is said to be integrable on (−∞, ∞) if it is integrable over (−∞, a) and (a, ∞) for some a ∈ R. In this case ∞ −∞ ∞ −∞ a f (x ) dx = −∞ f (x ) dx + a ∞ f (x ) dx . a Note that f (x ) dx may diverge even though lim a→∞ −a f (x ) dx exists. Improper Integrals 47 Improper Integrals 48 The p-integral Example 30 Show that ∞ −∞ a Theorem 31 x dx exists. The improper integral p ≤ 1. Proof: Consider first the case p = 1. Then, for some a > 1, a 1 1 ∞ x dx diverges but lim a→∞ −a 1 dx is convergent if p > 1 and divergent if xp 1 dx = ln x |a = ln a. 1 x Hence 1 a 1 dx −→ ∞ x 1 as a → ∞. ∞ Therefore, the improper integral Improper Integrals 49 Improper Integrals 1 dx diverges. x 50 Consider next the case p = 1. Then a 1 The inequality form of the comparison test 1 a 1− p − 1 . 1−p Theorem 32 Suppose that 0 ≤ f (x ) ≤ g (x ) ∀x ≥ a, and that f and g are integrable over [a, b ] for all b ≥ a. Then 1 1 x 1− p dx = xp 1−p a1−p = a = 1 If p > 1 then 1 −→ 0 as a → ∞. ap−1 a 1 Hence 1 ∞ 1 dx = lim a→∞ xp If p < 1 then Hence 1 1 1 . dx = xp p−1 If a ∞ ∞ g (x ) dx converges then f (x ) dx diverges then a a ∞ ∞ f (x ) dx converges. a1−p −→ ∞ as a → ∞. a 2 If a g (x ) dx diverges. 1 dx −→ ∞ xp ∞ 1 as a → ∞, i.e., the improper integral Improper Integrals 1 dx diverges. xp 51 Improper Integrals 52 Example 33 Determine if the following improper integrals converge. ∞ x2 1 √ dx x7 + 1 2 ∞ 1 2 dx 3/2 − 1 x 2 ∞ 1 3 dx x 1/2 + 1 2 Improper Integrals 53 Improper Integrals 54 The limit form of the comparison test Theorem 34 Suppose that f and g are nonnegative and bounded functions on [a, ∞), and that f (x ) = L. lim x →∞ g (x ) 1 Example 35 Determine if the following improper integrals converge. 1 ∞ 1 ∞ 1 e− √ x dx 2 sin x + 3/2 √ dx x +2 If 0 < L < ∞ then either ∞ ∞ a f (x ) dx and a ∞ g (x ) dx both 2 converge or they both diverge. If L = 0 and g (x ) dx converges, then so does a ∞ a ∞ ∞ f (x ) dx ; g (x ) dx . f (x ) dx diverges, then so does a a Improper Integrals 55 Improper Integrals 56 Non-elementary functions Many practically important functions are defined in terms of indefinite integrals. A few examples are: 2 erf(x ) = √ π x x 0 e−t dt 2 the error function the sine integral a Fresnel integral sin t dt t 0 x π2 t dt cos C(x ) = 2 0 S( x ) = All these integrals cannot be expressed as finite combinations of elementary functions. See the book Abramowitz and Stegun, “Handbook of Mathematical functions . . . ”, now online at http://www.iopb.res.in/~somen/abramowitz_and_stegun/ Improper Integrals 57 ...
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