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# 5.1 anti derivatives

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### 5.1 anti derivatives

1. 1. Antiderivatives
2. 2. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement..
3. 3. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function.
4. 4. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof.
5. 5. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k.
6. 6. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k. We are given that f'(x) = 0 so f(x) is differentiable.
7. 7. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k. We are given that f'(x) = 0 so f(x) is differentiable. By the Mean Value Theorem, there is a “c” where 0 < c < x and that f(x) – f(0) = f '(c) x – 0
8. 8. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k. We are given that f'(x) = 0 so f(x) is differentiable. By the Mean Value Theorem, there is a “c” where 0 < c < x and that f(x) – f(0) = f '(c) = 0 by the assumption, x – 0
9. 9. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k. We are given that f'(x) = 0 so f(x) is differentiable. By the Mean Value Theorem, there is a “c” where 0 < c < x and that f(x) – f(0) = f '(c) = 0 by the assumption, x – 0 hence f(x) – f(0) = 0, or that f(x) = f(0) = k.
10. 10. Antiderivatives The derivative of a constant function f(x) = k is 0. The Mean Value Theorem gives us the converse of the above statement. Theorem. If f'(x) = 0 for all x’s then f(x) is a constant function. Proof. We will show that for any x > 0 that f(x) = f(0) = k. We are given that f'(x) = 0 so f(x) is differentiable. By the Mean Value Theorem, there is a “c” where 0 < c < x and that f(x) – f(0) = f '(c) = 0 by the assumption, x – 0 hence f(x) – f(0) = 0, or that f(x) = f(0) = k. A similar argument may be made for x < 0, hence f(x) = k.
11. 11. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x).
12. 12. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) “the (indefinite) integral of f(x)”.
13. 13. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) “the (indefinite) integral of f(x)”. We write this as F(x) = ∫f(x) dx
14. 14. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) “the (indefinite) integral of f(x)”. We write this as ∫f(x) dx “The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k” or ∫0 dx = k F(x) = Hence the above theorem may be stated as: '
15. 15. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) “the (indefinite) integral of f(x)”. We write this as ∫f(x) dx Hence the above theorem may be stated as: “The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k” or ∫0 dx = k F(x) = ' Remarks I. There is only one derivative for any f(x) but the “antiderivatives” of f(x) is a set of functions.
16. 16. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) “the (indefinite) integral of f(x)”. We write this as ∫f(x) dx Hence the above theorem may be stated as: “The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k” or ∫0 dx = k F(x) = ' Remarks I. There is only one derivative for any f(x) but the “antiderivatives” of f(x) is a set of functions. II. Any two antiderivatives of f(x) must differ by a constant. Hence if F'(x) = G'(x) then F(x) = G(x) + k for some constant k.
17. 17. Antiderivatives The procedure of finding the antiderivatives is called “integration”.
18. 18. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules
19. 19. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g.
20. 20. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx
21. 21. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”.
22. 22. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”. For any constant c, (c*F) ' = c*F'= c*f,
23. 23. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, ∫cf dx = c ∫f dx
24. 24. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, ∫cf dx = c ∫f dx We say that we can “pull out the constant multiple.”
25. 25. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, ∫cf dx = c ∫f dx We say that we can “pull out the constant multiple.” The following is a list of integration formulas for basic functions.
26. 26. Antiderivatives The procedure of finding the antiderivatives is called “integration”. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules Let f, g, F, G and that F' = f and G' = g, then (F ± G)' = F' ± G' = f + g. In integral notation, ∫f ± g dx = ∫f dx ±∫ g dx and we say that we can “integrate term by term”. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, ∫cf dx = c ∫f dx We say that we can “pull out the constant multiple.” The following is a list of integration formulas for basic functions. These formulas are just the derivatives of the basic functions written in the integral forms.
27. 27. ∫ xpdx xP+1 = Antiderivatives sin(x) dx = –cos(x) + k sec(x)tan(x) dx ∫ ex dx = ex + k csc2(x) dx 1 x dx = ln(x) + k ∫ ∫ P + 1 + k where P ≠ –1 ∫ cos(x) dx = sin(x) + k ∫sec2(x) dx = tan(x) + k ∫ = –cot(x) + k ∫ = sec(x) + k ∫csc(x)cot(x) dx = –csc(x) + k The Power Functions The Trig–Functions The Log and Exponential Functions ∫ x–1 dx = ln(x) + k or
28. 28. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx
29. 29. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx Integrate term by term and pull out the constant multiple of each term.
30. 30. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x = 3 ∫ ex dx dx Integrate term by term and pull out the constant multiple of each term.
31. 31. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x = 3 ∫ ex dx – 5 ∫ sin(x) dx dx Integrate term by term and pull out the constant multiple of each term.
32. 32. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx Integrate term by term and pull out the constant multiple of each term.
33. 33. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx Integrate term by term and pull out the constant multiple of each term. There is no Quotient Rule for integration. Hence we have to separate the fractions.
34. 34. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k
35. 35. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k Although each integral yields a constant term, these terms may be collected as a single “k”.
36. 36. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k The graphs of all the antiderivatives F(x) + k of f(x) are positioned by the integration constant k.
37. 37. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k The graphs of all the antiderivatives F(x) + k of f(x) are positioned by the integration constant k. For example, the antiderivatives of 2x are F(x) = x2 + k.
38. 38. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k The graphs of all the antiderivatives F(x) + k of f(x) are positioned by the k=2 integration constant k. k=1 For example, the antiderivatives of 2x are F(x) = x2 + k. Their graphs k=0 are just the vertical translations of k=–1 the parabola y = x2 “filed” by k as k=–2 shown here. Graphs of the antiderivatives of 2x, F(x) = x2 + k, layered by k
39. 39. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx x 2 = 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ dx = 3ex + 5cos(x) + 2x½ – 2 ln(x) + k The graphs of all the antiderivatives F(x) + k of f(x) are positioned by the integration constant k. For example, the antiderivatives of 2x are F(x) = x2 + k. Their graphs are just the vertical translations of the parabola y = x2 “filed” by k as shown here. All of them have identical slope at any given x. k=2 k=1 k=0 k=–1 k=–2 Graphs of the antiderivatives of 2x, F(x) = x2 + k, layered by k
40. 40. Antiderivatives When solving for the equation of a line with a given derivative, i.e. its slope m, a point on the line is needed in order to determine its equation. (1, 3) k=2 k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k
41. 41. Antiderivatives When solving for the equation of a line with a given derivative, i.e. its slope m, a point on the line is needed in order to determine its equation. Likewise if we know the derivative, i.e. the slopes, of a function F(x) then all we need is a point to completely determine F(x). (1, 3) k=2 k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k
42. 42. Antiderivatives When solving for the equation of a line with a given derivative, i.e. its slope m, a point on the line is needed in order to determine its equation. Likewise if we know the derivative, i.e. the slopes, of a function F(x) then all we need is a point to completely determine F(x). k=2 k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k Such a data–point is called the “initial condition”. (1, 3)
43. 43. Antiderivatives When solving for the equation of a line with a given derivative, i.e. its slope m, a point on the line is needed in order to determine its equation. Likewise if we know the derivative, i.e. the slopes, of a function F(x) then all we need is a point to completely determine F(x). k=2 k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k Such a data–point is called the “initial condition”. In general the initial condition is a list of data– points that enables us to determine all the unspecified constants in the general solution. (1, 3)
44. 44. Antiderivatives When solving for the equation of a line with a given derivative, i.e. its slope m, a point on the line is needed in order to determine its equation. Likewise if we know the derivative, i.e. the slopes, of a function F(x) then all we need is a point to completely determine F(x). k=2 k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k Such a data–point is called the “initial condition”. In general the initial condition is a list of data– points that enables us to determine all the unspecified constants in the general solution. So if F'(x) = 2x with the initial condition F(1) = 3, then it must be that F(x) = x2 + 2. (1, 3)
45. 45. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information.
46. 46. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 t 1 with the initial condition x'(1) = 1/3 and that x(1) = 1/12.
47. 47. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 t 1 with the initial condition x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t).
48. 48. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 with the initial condition t 1 dt So ∫ t 1 x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t). x'(t) = (t – )2
49. 49. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 ∫ t 1 x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t). t 1 dt x'(t) = (t – )2 = ∫ t2 – 2 + 1 dt t2 So with the initial condition
50. 50. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 ∫ t 1 with the initial condition x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t). t 1 dt x'(t) = (t – )2 = ∫ t2 – 2 + 1 dt t2 = t3 3 – 2t – t–1 + k So where k is a constant.
51. 51. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 ∫ t 1 with the initial condition x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t). t 1 dt x'(t) = (t – )2 = ∫ t2 – 2 + 1 dt t2 = t3 3 – 2t – t–1 + k So where k is a constant. We are given that x'(1) = 1/3, so 1 = 1/3 3 – 2 – 1 + k
52. 52. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information. Example B. Find the position function x(t) given that x''(t) = (t – )2 ∫ t 1 with the initial condition x'(1) = 1/3 and that x(1) = 1/12. The 1st derivative, x'(t) is the integral of x''(t). t 1 dt x'(t) = (t – )2 = ∫ t2 – 2 + 1 dt t2 = t3 3 – 2t – t–1 + k So where k is a constant. We are given that x'(1) = 1/3, so 1 = 1/3 3 – 2 – 1 + k Therefore k = 3 and that t3 x'(t) = – 2t – t–1 + 3 3
53. 53. Antiderivatives The function x(t) is an antiderivative of x'(t).
54. 54. Antiderivatives The function x(t) is an antiderivative of x'(t). So x(t) = ∫ t dt 3 3 – 2t – t–1 + 3
55. 55. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K where K is a constant.
56. 56. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so
57. 57. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/12 12 where K is a constant.
58. 58. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/12 12 We have that K = –2. where K is a constant.
59. 59. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/12 12 We have that K = –2. Therefore x(t) = t4 12 – t2 – ln(t) + 3t – 2
60. 60. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/12 12 We have that K = –2. Therefore x(t) = t4 12 – t2 – ln(t) + 3t – 2 As we noticed before, the formulas whose higher order derivatives eventually become 0 are the polynomials.
61. 61. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = ∫ t – 2t – t–1 + 3 dt 3 = t4 12 – t2 – ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/12 12 We have that K = –2. Therefore x(t) = t4 12 – t2 – ln(t) + 3t – 2 As we noticed before, the formulas whose higher order derivatives eventually become 0 are the polynomials. Specifically if P(x) is a polynomial of deg N, then P(N+1) (x) = P(N+2) (x) = P(N+3) (x) = .. = 0.
62. 62. Antiderivatives Theorem. If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for some polynomial P(x) with deg(P) < N.
63. 63. Antiderivatives Theorem. If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for some polynomial P(x) with deg(P) < N. In other words, if two function have the same N’th derivatives, then their difference is a polynomial with degree less than N.
64. 64. Antiderivatives Theorem. If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for some polynomial P(x) with deg(P) < N. In other words, if two function have the same N’th derivatives, then their difference is a polynomial with degree less than N. In general, we need N data points in the initial condition to recover the difference P(x) completely.