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5.1 anti derivatives

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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&#x2019;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&#x2019;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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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 &#x201C;c&#x201D; where 0 &lt; c &lt; x and that f(x) &#x2013; f(0) = f '(c) x &#x2013; 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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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 &#x201C;c&#x201D; where 0 &lt; c &lt; x and that f(x) &#x2013; f(0) = f '(c) = 0 by the assumption, x &#x2013; 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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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 &#x201C;c&#x201D; where 0 &lt; c &lt; x and that f(x) &#x2013; f(0) = f '(c) = 0 by the assumption, x &#x2013; 0 hence f(x) &#x2013; 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&#x2019;s then f(x) is a constant function. Proof. We will show that for any x &gt; 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 &#x201C;c&#x201D; where 0 &lt; c &lt; x and that f(x) &#x2013; f(0) = f '(c) = 0 by the assumption, x &#x2013; 0 hence f(x) &#x2013; f(0) = 0, or that f(x) = f(0) = k. A similar argument may be made for x &lt; 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) &#x201C;the (indefinite) integral of f(x)&#x201D;.
• 13. Antiderivatives We say F(x) is an antiderivative of f(x) if F'(x) = f(x). We also call F(x) &#x201C;the (indefinite) integral of f(x)&#x201D;. We write this as F(x) = &#x222B;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) &#x201C;the (indefinite) integral of f(x)&#x201D;. We write this as &#x222B;f(x) dx &#x201C;The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k&#x201D; or &#x222B;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) &#x201C;the (indefinite) integral of f(x)&#x201D;. We write this as &#x222B;f(x) dx Hence the above theorem may be stated as: &#x201C;The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k&#x201D; or &#x222B;0 dx = k F(x) = ' Remarks I. There is only one derivative for any f(x) but the &#x201C;antiderivatives&#x201D; 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) &#x201C;the (indefinite) integral of f(x)&#x201D;. We write this as &#x222B;f(x) dx Hence the above theorem may be stated as: &#x201C;The antiderivatives of the function f(x) = 0 are exactly all the constant functions F(x) = k&#x201D; or &#x222B;0 dx = k F(x) = ' Remarks I. There is only one derivative for any f(x) but the &#x201C;antiderivatives&#x201D; 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 &#x201C;integration&#x201D;.
• 18. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. Rewording the basic differentiation rules, we have the following integration rules. Basic Integration Rules
• 19. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g.
• 20. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx
• 21. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;.
• 22. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;. For any constant c, (c*F) ' = c*F'= c*f,
• 23. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, &#x222B;cf dx = c &#x222B;f dx
• 24. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, &#x222B;cf dx = c &#x222B;f dx We say that we can &#x201C;pull out the constant multiple.&#x201D;
• 25. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, &#x222B;cf dx = c &#x222B;f dx We say that we can &#x201C;pull out the constant multiple.&#x201D; The following is a list of integration formulas for basic functions.
• 26. Antiderivatives The procedure of finding the antiderivatives is called &#x201C;integration&#x201D;. 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 &#xB1; G)' = F' &#xB1; G' = f + g. In integral notation, &#x222B;f &#xB1; g dx = &#x222B;f dx &#xB1;&#x222B; g dx and we say that we can &#x201C;integrate term by term&#x201D;. For any constant c, (c*F) ' = c*F'= c*f, so in integrals, &#x222B;cf dx = c &#x222B;f dx We say that we can &#x201C;pull out the constant multiple.&#x201D; 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. &#x222B; xpdx xP+1 = Antiderivatives sin(x) dx = &#x2013;cos(x) + k sec(x)tan(x) dx &#x222B; ex dx = ex + k csc2(x) dx 1 x dx = ln(x) + k &#x222B; &#x222B; P + 1 + k where P &#x2260; &#x2013;1 &#x222B; cos(x) dx = sin(x) + k &#x222B;sec2(x) dx = tan(x) + k &#x222B; = &#x2013;cot(x) + k &#x222B; = sec(x) + k &#x222B;csc(x)cot(x) dx = &#x2013;csc(x) + k The Power Functions The Trig&#x2013;Functions The Log and Exponential Functions &#x222B; x&#x2013;1 dx = ln(x) + k or
• 28. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx
• 29. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx Integrate term by term and pull out the constant multiple of each term.
• 30. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x = 3 &#x222B; ex dx dx Integrate term by term and pull out the constant multiple of each term.
• 31. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x = 3 &#x222B; ex dx &#x2013; 5 &#x222B; 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. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx Integrate term by term and pull out the constant multiple of each term.
• 33. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; 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. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 2 ln(x) + k
• 35. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 2 ln(x) + k Although each integral yields a constant term, these terms may be collected as a single &#x201C;k&#x201D;.
• 36. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 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. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 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. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 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=&#x2013;1 the parabola y = x2 &#x201C;filed&#x201D; by k as k=&#x2013;2 shown here. Graphs of the antiderivatives of 2x, F(x) = x2 + k, layered by k
• 39. Antiderivatives Example A. Find the following antiderivative. &#x222B;3ex &#x2013; 5sin(x) + &#x221A;x &#x2013; 2 x dx x 2 = 3 &#x222B; ex dx &#x2013; 5 &#x222B; sin(x) dx + &#x222B; x&#x2013;&#xBD; dx &#x2013; &#x222B; dx = 3ex + 5cos(x) + 2x&#xBD; &#x2013; 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 &#x201C;filed&#x201D; by k as shown here. All of them have identical slope at any given x. k=2 k=1 k=0 k=&#x2013;1 k=&#x2013;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=&#x2013;1 k=&#x2013;2 antiderivatives of 2x, F(x) = x2 + k, &#x201C;filed&#x201D; 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=&#x2013;1 k=&#x2013;2 antiderivatives of 2x, F(x) = x2 + k, &#x201C;filed&#x201D; 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=&#x2013;1 k=&#x2013;2 antiderivatives of 2x, F(x) = x2 + k, &#x201C;filed&#x201D; by k Such a data&#x2013;point is called the &#x201C;initial condition&#x201D;. (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=&#x2013;1 k=&#x2013;2 antiderivatives of 2x, F(x) = x2 + k, &#x201C;filed&#x201D; by k Such a data&#x2013;point is called the &#x201C;initial condition&#x201D;. In general the initial condition is a list of data&#x2013; 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=&#x2013;1 k=&#x2013;2 antiderivatives of 2x, F(x) = x2 + k, &#x201C;filed&#x201D; by k Such a data&#x2013;point is called the &#x201C;initial condition&#x201D;. In general the initial condition is a list of data&#x2013; 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 &#x2013; )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 &#x2013; )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 &#x2013; )2 with the initial condition t 1 dt So &#x222B; 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 &#x2013; )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 &#x2013; )2 &#x222B; 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 &#x2013; )2 = &#x222B; t2 &#x2013; 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 &#x2013; )2 &#x222B; 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 &#x2013; )2 = &#x222B; t2 &#x2013; 2 + 1 dt t2 = t3 3 &#x2013; 2t &#x2013; t&#x2013;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 &#x2013; )2 &#x222B; 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 &#x2013; )2 = &#x222B; t2 &#x2013; 2 + 1 dt t2 = t3 3 &#x2013; 2t &#x2013; t&#x2013;1 + k So where k is a constant. We are given that x'(1) = 1/3, so 1 = 1/3 3 &#x2013; 2 &#x2013; 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 &#x2013; )2 &#x222B; 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 &#x2013; )2 = &#x222B; t2 &#x2013; 2 + 1 dt t2 = t3 3 &#x2013; 2t &#x2013; t&#x2013;1 + k So where k is a constant. We are given that x'(1) = 1/3, so 1 = 1/3 3 &#x2013; 2 &#x2013; 1 + k Therefore k = 3 and that t3 x'(t) = &#x2013; 2t &#x2013; t&#x2013;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) = &#x222B; t dt 3 3 &#x2013; 2t &#x2013; t&#x2013;1 + 3
• 55. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; 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) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; 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) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t + K We are given that x(1) = 1/12, so 1 &#x2013; 1 &#x2013; 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) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t + K We are given that x(1) = 1/12, so 1 &#x2013; 1 &#x2013; 0 + 3 + K = 1/12 12 We have that K = &#x2013;2. where K is a constant.
• 59. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 &#x2013; 1 &#x2013; 0 + 3 + K = 1/12 12 We have that K = &#x2013;2. Therefore x(t) = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t &#x2013; 2
• 60. Antiderivatives The function x(t) is an antiderivative of x'(t). 3 So x(t) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 &#x2013; 1 &#x2013; 0 + 3 + K = 1/12 12 We have that K = &#x2013;2. Therefore x(t) = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t &#x2013; 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) = &#x222B; t &#x2013; 2t &#x2013; t&#x2013;1 + 3 dt 3 = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t + K where K is a constant. We are given that x(1) = 1/12, so 1 &#x2013; 1 &#x2013; 0 + 3 + K = 1/12 12 We have that K = &#x2013;2. Therefore x(t) = t4 12 &#x2013; t2 &#x2013; ln(t) + 3t &#x2013; 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) &lt; 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) &lt; N. In other words, if two function have the same N&#x2019;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) &lt; N. In other words, if two function have the same N&#x2019;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.