• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
5.1 anti derivatives
 

5.1 anti derivatives

on

  • 543 views

 

Statistics

Views

Total Views
543
Views on SlideShare
543
Embed Views
0

Actions

Likes
0
Downloads
0
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    5.1 anti derivatives 5.1 anti derivatives Presentation Transcript

    • Antiderivatives
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statement
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.If f(x) = 0 for all x’s then f(x) is a constant function.
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.If f(x) = 0 for all x’s then f(x) is a constant function.Proof.
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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.
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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.
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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” where0 < c < x and thatf(x) – f(0) = f (c) x–0
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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” where0 < c < x and thatf(x) – f(0) = f (c) = 0 by the assumption, x–0
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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” where0 < c < x and thatf(x) – f(0) = f (c) = 0 by the assumption, x–0hence f(x) – f(0) = 0, or that f(x) = f(0) = k.
    • AntiderivativesThe derivative of a constant function f(x) = k is 0.The Mean Value Theorem gives us the converse ofthe above statementTheorem.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” where0 < c < x and thatf(x) – f(0) = f (c) = 0 by the assumption, x–0hence f(x) – f(0) = 0, or that f(x) = f(0) = k. A similarargument may be made for x < 0, hence f(x) = k.
    • AntiderivativesWe say F(x) is an antiderivative of f(x) if F(x) = f(x).
    • AntiderivativesWe 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)”.
    • AntiderivativesWe 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
    • 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) dxHence 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
    • 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) dxHence 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 = kRemarksI. There is only one derivative for any f(x) but the“antiderivatives” of f(x) is a set of functions.
    • 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) dxHence 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 = kRemarksI. 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 aconstant. Hence if F(x) = G(x) then F(x) = G(x) + kfor some constant k.
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”.
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration Rules
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet f, g, F, G and that F = f and G = g,then (F ± G) = F ± G = f + g.
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand we say that we can “integrate term by term”.
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand we say that we can “integrate term by term”.For any constant c, (c*F) = c*F= c*f,
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand 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
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand 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 dxWe say that we can “pull out the constant multiple.”
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand 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 dxWe say that we can “pull out the constant multiple.”The following is a list of integration formulas for basicfunctions.
    • AntiderivativesThe procedure of finding the antiderivatives is called“integration”. Rewording the basic differentiationrules, we have the following integration rules.Basic Integration RulesLet 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 dxand 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 dxWe say that we can “pull out the constant multiple.”The following is a list of integration formulas for basicfunctions. These formulas are just the derivatives ofthe basic functions written in the integral forms.
    • AntiderivativesThe Power Functions∫x p dx = xP+1 P + 1 + k where P ≠ –1The Trig–Functions∫ sin(x) dx = –cos(x) + k ∫ cos(x) dx = sin(x) + k∫sec2(x) dx = tan(x) + k ∫ csc2(x) dx = –cot(x) + k∫ sec(x)tan(x) dx ∫ csc(x)cot(x) dx = sec(x) + k = –csc(x) + kThe Log and Exponential Functions ∫x1 dx = ln(x) + k or∫e x dx = ex + k ∫ x–1 dx = ln(x) + k
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx – Integrate term by term and pull out the constant multiple of each term.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx Integrate term by term and pull out the constant multiple of each term.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx Integrate term by term and pull out the constant multiple of each term.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x Integrate term by term and pull out the constant multiple of each term.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x Integrate term by term and pull out the constant There is no Quotient Rule multiple of each term. for integration. Hence we have to separate the fractions.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + k
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + k Although each integral yields a constant term, these terms may be collected as a single “k”.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + kThe graphs of all the antiderivativesF(x) + k of f(x) are positioned by theintegration constant k.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + kThe graphs of all the antiderivativesF(x) + k of f(x) are positioned by theintegration constant k.For example, the antiderivatives of2x are F(x) = x2 + k.
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + kThe graphs of all the antiderivativesF(x) + k of f(x) are positioned by the k=2integration constant k. k=1For example, the antiderivatives of k=02x are F(x) = x2 + k. Their graphsare just the vertical translations of k=–1the parabola y = x2 “filed” by k as k=–2shown here. Graphs of the antiderivatives of 2x, F(x) = x2 + k, layered by k
    • AntiderivativesExample A. Find the following antiderivative. ∫ 3ex – 5sin(x) + √x x 2 dx –= 3 ∫ ex dx – 5 ∫ sin(x) dx + ∫ x–½ dx – ∫ 2 dx x= 3ex + 5cos(x) + 2x½ – 2 ln(x) + kThe graphs of all the antiderivativesF(x) + k of f(x) are positioned by the k=2integration constant k. k=1For example, the antiderivatives of k=02x are F(x) = x2 + k. Their graphsare just the vertical translations of k=–1the parabola y = x2 “filed” by k as k=–2shown here. All of them have Graphs of the antiderivativesidentical slope at any given x. of 2x, F(x) = x2 + k, layered by k
    • AntiderivativesWhen solving for the equation of a line with a givenderivative, i.e. its slope m, a point on the line isneeded 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
    • AntiderivativesWhen solving for the equation of a line with a givenderivative, i.e. its slope m, a point on the line isneeded 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 tocompletely 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
    • AntiderivativesWhen solving for the equation of a line with a givenderivative, i.e. its slope m, a point on the line isneeded 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 tocompletely determine F(x). (1, 3)Such a data–point is called the k=2“initial condition”. k=1 k=0 k=–1 k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k
    • AntiderivativesWhen solving for the equation of a line with a givenderivative, i.e. its slope m, a point on the line isneeded 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 tocompletely determine F(x). (1, 3)Such a data–point is called the k=2“initial condition”. In general theinitial condition is a list of data– k=1points that enables us to determine k=0all the unspecified constants in the k=–1general solution. k=–2 antiderivatives of 2x, F(x) = x2 + k, “filed” by k
    • AntiderivativesWhen solving for the equation of a line with a givenderivative, i.e. its slope m, a point on the line isneeded 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 tocompletely determine F(x). (1, 3)Such a data–point is called the k=2“initial condition”. In general theinitial condition is a list of data– k=1points that enables us to determine k=0all the unspecified constants in the k=–1general solution. So if F(x) = 2xwith the initial condition F(1) = 3, k=–2then it must be that F(x) = x2 + 2. antiderivatives of 2x, F(x) = x + k, “filed” by k 2
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).So x(t) = ∫ (t – 1)2 dt t
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).So x(t) = ∫ (t – 1)2 dt t = ∫ t2 – 2 + 1 dt t2
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).So x(t) = ∫ (t – 1)2 dt t = ∫ t2 – 2 + 1 dt t2 3 = t – 2t – t–1 + k where k is a constant. 3
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).So x(t) = ∫ (t – 1)2 dt t =∫ t2 – 2 + 1 dt t2 3 = t – 2t – t–1 + k where k is a constant. 3We are given that x(1) = 1/3, so 1 – 2 – 1 + k = 1/3 3
    • AntiderivativesWe may recover a function completely through itshigher order derivative if we have sufficient information.Example B. Find the position function x(t) given that 1x(t) = (t – t )2 with the initial conditionx(1) = 1/3 and that x(1) = 1/12.The 1st derivative x(t) is the integral of the x(t).So x(t) = ∫ (t – 1)2 dt t =∫ t2 – 2 + 1 dt t2 3 = t – 2t – t–1 + k where k is a constant. 3We are given that x(1) = 1/3, so 1 – 2 – 1 + k = 1/3 3 t3 – 2t – t–1 + 3Therefore k = 3 and that x(t) = 3
    • AntiderivativesThe function x(t) is an antiderivative of x(t).
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/1212
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/1212We have that K = –2.
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/1212We have that K = –2.Therefore x(t) = t4 – t2 – ln(t) + 3t – 2 12
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/1212We have that K = –2.Therefore x(t) = t4 – t2 – ln(t) + 3t – 2 12As we noticed before, the formulas whose higherorder derivatives eventually become 0 are thepolynomials.
    • AntiderivativesThe function x(t) is an antiderivative of x(t). t3 – 2t – t–1 + 3 dtSo x(t) = ∫ 3 t4 – t2 – ln(t) + 3t + K where K is a constant.= 12We are given that x(1) = 1/12, so 1 – 1 – 0 + 3 + K = 1/1212We have that K = –2.Therefore x(t) = t4 – t2 – ln(t) + 3t – 2 12As we noticed before, the formulas whose higherorder derivatives eventually become 0 are thepolynomials. Specifically if P(x) is a polynomial ofdeg N, then P(N+1) (x) = P(N+2) (x) = P(N+3) (x) = .. = 0.
    • AntiderivativesTheorem.If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for somepolynomial P(x) with deg(P) < N.
    • AntiderivativesTheorem.If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for somepolynomial P(x) with deg(P) < N.In other words, if two function have the same N’thderivatives, then their difference is a polynomial withdegree less than N.
    • AntiderivativesTheorem.If F(N) (x) = G(N)(x) then F(x) = G(x) + P(x) for somepolynomial P(x) with deg(P) < N.In other words, if two function have the same N’thderivatives, then their difference is a polynomial withdegree less than N. In general, we need N data pointsin the initial condition to recover the differenceP(x) completely.