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- 1. Antiderivatives
- 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. 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. 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. 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. 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. 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. 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. 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. 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. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x).
- 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. 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. 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. 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. 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. Antiderivatives The procedure of finding the antiderivatives is called “integration”.
- 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. 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. 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. 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. 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. 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. 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. 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. 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. ∫ 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. Antiderivatives Example A. Find the following antiderivative. ∫3ex – 5sin(x) + √x – 2 x dx
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Antiderivatives We may recover a function completely through its higher order derivative if we have sufficient information.
- 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. 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. 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. 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. 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. 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. 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. Antiderivatives The function x(t) is an antiderivative of x'(t).
- 54. Antiderivatives The function x(t) is an antiderivative of x'(t). So x(t) = ∫ t dt 3 3 – 2t – t–1 + 3
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.

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