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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