反微分與不定積分
及其性質
1.
2.
3.
4.
反微分(Antiderivatives)
不定積分(Indefinite Integral)
積分規則(Rules of Integration)
替代法(Substitution)
page 358~379
Definition of Antiderivative:
If F’(x)=f(x), then F(x) is an
antiderivative of f(x)
Example
|
EX1(a)
F(x)=10x, F’(x)=10
Æ F(x) is an antiderivative of f(x)=10
Example
|
EX1(b)
2
x
F(x)= , F’(x)=2x
Æ F(x) is an antiderivative of f(x)=2x
Example
|
4
5x
EX2 Find an antiderivative of f(x)=
g ( x) = x n , g ′( x) = nx n−1
Æ g ′( x) = nx
n −1
is an antiderivative of g ( x) = x
n
4
When n=5, x is an antiderivative of 5x
1 n+1
n
Æ An antiderivative of x is n + 1 x
n
Example
|
EX3 Find an antiderivative
ofx f(x)=
x
e
e
let F(x)= and F’(x)= =f(x)
ex
Æ F(x) is an Antiderivative of f(x)
Notice
F(x)= e x
and F’(x)=e x
let G(x)= e x +2 and G’(x)= e x =f(x)
x
e
let H(x)= e +100 and H’(x)= =f(x)
x
Æ
F(x), G(x) and H(x) are antiderivative of f(x)
x
e
Antiderivatives of f(x) differ by a constant
Property of Antiderivatives
If F(x) and G(x) are both antiderivatives of a
function f(x) on an interval, then there is a
constant C such that
F(x)-G(x)=C
The arbitrary real number C is called an
integration constant (積分常數)
EX: F(x)=2x+2 and F’(x)=2
G(x)=2x+100 and G’(x)=2
H(x)=2x+10000 and H’(x)=2
Indefinite Integral
|
The family of all antiderivative of f(x) (F’(x)
= f(x)) is indicated by
∫ f ( x)dx = F ( x) + C
where C : Integral Constant
∫ : Integral Sign(積分符號)
f(x) : Integrand(積分函數)
dx : integral of f(x) with respect to x
|
∫ f ( x)dx is called an Indefinite Integral
Power Rule
|
For any real number n ≠ −1,
x dx =
+C
∫
EX4 Use the power rule to find each indefinite integral.
n
(a)
(b)
(c)
(d)
x n +1
n +1
t 3 +1
t4
∫ t dt = 3 + 1 + C = 4 + C
− 2 +1
1
t
−2
dt
t
=
∫ t2
∫ dt = − 2 + 1 + C =
1
1 2 +1
t
2
t
dt
t
=
∫
∫ dt = 1 2 + 1 + C =
t 0 +1
0
∫ 1dt = ∫ t dt = 0 + 1 + C = t +
3
−
1
+C
t
2 32
t +C
3
C
Constant Multiple Rule and
Sum or Difference Rule
|
If all indicated integrals exist,
∫ a ⋅ f ( x)dx = a ∫ f ( x)dx
and
∫ [ af ( x) ± bg ( x)] dx = a∫ f ( x)dx ± b∫ g ( x)dx
for any real number a, b
Example: Find
f ( x) = x
4/3
→ ∫ x 4 / 3 dx = 73 x 7 / 3 +C
Example: Find
2
(3x
∫ + 4x)dx
2
2
(
3
x
+
4
x
)
dx
=
3
x
∫
∫ dx + 4∫ xdx
2
x3
x2
x
= 3( +C 1) + 4( + C2 ) = x 3 +
+C
3
2
2
because C1,C2 are constants. Then we
can use C to represent integral constant
where C = C 1 +C2
EX5 Use the rules to find each integral.
(a)
3
3
2
ν
d
ν
=
2
ν
∫
∫ dν
(by constant multiple rule)
ν3+1
1 4
=2
+C = ν +C (by power rule)
3+1
2
12
(b) ∫ 5 dz = 12 ∫ z −5 dz
z
z − 5 +1
= 12
+ C = − 3 z −4 + C
−5 + 1
EX6 Use the rules to find each integral
(c)
4
2
x
−
1
dx
=
x
−
2
x
+ 1) dx
)
∫(
∫(
2
2
x 4+1
x 2+1
=
−2
+ x+C
4 +1
2 +1
EX5 Use the rules to find each integral.
(a)
3
3
2
ν
d
ν
=
2
ν
∫
∫ dν
(by constant multiple rule)
ν3+1
1 4
=2
+C = ν +C (by power rule)
3+1
2
12
(b) ∫ 5 dz = 12 ∫ z −5 dz
z
z − 5 +1
= 12
+ C = − 3 z −4 + C
−5 + 1
Review of Derivative of
Exponential Function
|
x
x
e
e
f(x) =
Æ f’(x) =
|
f(x) = a = e
|
kx
kx
e
f(x) =
Æ f’(x) = ke
|
kx
kx ln a
a
=
e
f(x) =
Æ f’(x) = k ( ln a ) a kx
x
x ln a
x
ln
a
a
)
Æ f’(x) = (
Indefinite Integrals of
Exponential Functions
|
∫
e xdx = e x + C
kx
|
|
|
e
∫ e dx = k + C , k ≠ 0
x
a
x
a
∫ d x = ln a + C
kx
∫
kx
a
+ C , k ≠0
a kx d x =
k ( ln a )
Review of Derivative of
Exponential Function
|
x
x
e
e
f(x) =
Æ f’(x) =
|
f(x) = a = e
|
kx
kx
e
f(x) =
Æ f’(x) = ke
|
kx
kx ln a
a
=
e
f(x) =
Æ f’(x) = k ( ln a ) a kx
x
x ln a
x
ln
a
a
)
Æ f’(x) = (
Indefinite Integrals of
Exponential Functions
|
∫
e xdx = e x + C
kx
|
|
|
e
∫ e dx = k + C , k ≠ 0
x
a
x
a
∫ d x = ln a + C
kx
∫
kx
a
+ C , k ≠0
a kx d x =
k ( ln a )
EX7 Exponential Functions
(a)
t
t
t
9
e
dt
=
9
e
dt
=
9
e
+C
∫
∫
9x
e
(b) ∫ e9 x dx =
+C
9
(c)
∫ 3e
5u
4
du = 3
e
5
4
5
4
u
1 2 54 u
e +C
+C =
5
−5 x
−5 x
2
2
−5 x
2
dx =
+C =−
+C
(d) ∫
5 ( ln 2 )
− 5 ( ln 2 )
Indefinite Integrals of
Exponential Functions
|
∫
e xdx = e x + C
kx
|
|
|
e
∫ e dx = k + C , k ≠ 0
x
a
x
a
∫ d x = ln a + C
kx
∫
kx
a
+ C , k ≠0
a kx d x =
k ( ln a )
EX7 Exponential Functions
(a)
t
t
t
9
e
dt
=
9
e
dt
=
9
e
+C
∫
∫
9x
e
(b) ∫ e9 x dx =
+C
9
(c)
∫ 3e
5u
4
du = 3
e
5
4
5
4
u
1 2 54 u
e +C
+C =
5
−5 x
−5 x
2
2
−5 x
2
dx =
+C =−
+C
(d) ∫
5 ( ln 2 )
− 5 ( ln 2 )
Indefinite Integral of
|
|
|
x
−1
Derivative of Logarithmic Function
f(x) = ln x where x ≠ 0
1
= x −1
Æ f’(x) = x
−1
1
x
dx
=
∫
∫ x dx = ln x + C
⎧ ln x , n = −1
⎪ n +1
n
∫ x dx = ⎨ x , n ≠ −1
⎪
⎩n +1
EX8 Integrals
(a)
4
1
∫ x dx = 4 ∫ x dx = 4 ln x + C
(b)
∫(−
5
x
+e
−2 x
) dx = −5ln x −
1
2
e
−2 x
+C
EX9 Cost-1
Suppose a publishing company has found that
the marginal cost at a level of production of x
thousand books is given by
50
C '( x) =
x
and that the fixed cost (the cost before the firsr
book can be produced) is $25,000. Find the cost
function
∫(−
5
x
+e
−2 x
) dx = −5ln x −
1
2
e
−2 x
+C
EX9 Cost-2
and use the indefinite integral rules to integrate the
50
− 12
function C '( x ) =
= 50 x
x
− 12 +1
1
x
2
50
x
dx
=
50
+
K
=
100
x
+K
∫
1
− 2 +1
− 12
When x=0, C(0)=25,000, K=25,000
The cost function is C ( x) = 100 x + 25,000
Review of the Chain Rule
d
⎡ f ( g ( x ) )⎤ = f ′ ( g ( x ) ) ⋅ g ′ ( x )
⎦
dx ⎣
du
1. Let u = g(x), dx =g’(x)
dw
2. Let w = f(g(x))=f(u), du =f’(u)
3.
dw
dx
=
dw
du
×
du
dx
Æ f(g(x)) = f’(u) g’(x)=f’(g(x))g’(x)
Substitution Rules
|
If u=g(x) is a differential function where
du=g’(x)dx, then
∫ F '(g(x))g '(x)dx = F(g(x)) + C
EX: ∫ 10 x ( x − 1) dx = ( x − 1 ) + C
2
4
2
5
1. u = x −1, du=2xdx
2
2
4
4
10
x
(
x
−
1)
dx
=
5
u
2. ∫
∫ du
= u 5 + C = ( x 2 + 1)5 + C
General Power Rule for Integrals
For u=f(x) and du=f’(x)dx,
n +1
u
n
u
∫ dx = n + 1 + C
2
4
6
x
(3
x
+
4)
dx
EX1: Find ∫
let u =3 x 2 + 4, du =6x
2
4
2
4
6
x
(3
x
+
4)
dx
=
(3
x
+
4)
(6 xdx)
Æ ∫
∫
5
1
u
= ∫ u 4 du = + C = (3 x 2 + 4)5 + C
5
5
EX2 General Power Rule
2
x
Find ∫
Let u=
∫
x 3 + 1d x
x + 1 , du = 3x 2dx
3
x + 1( x dx ) = ∫ u du =
3
2
1
2
u
1
2
1 +1
2
+1
+C
3
2 32
2 3
2
= u + C = ( x + 1) + C
3
3
EX3 General Power Rule
Find
∫
(x
x+3
2
+ 6x)
2
dx
Let u= x + 6 xdu
, = ( 2 x + 6 ) dx
1 2( x + 3)
1 −2
x+3
∫ x 2 + 6 x 2 dx = 2 ∫ ( x 2 + 6 x)2 dx = 2 ∫ u du
(
)
2
1 2
1 u −1
=
+C =
x + 6x
2 −1
2
(
)
−1
+C
Indefinite Integrals of
e
u
For u=f(x) and du=f’(x)dx,
∫ e dx = e
u
u
+C
Indefinite Integrals of
For u=f(x) and du=f’(x)dx,
du
∫ u dx = ∫ u = ln u + C
−1
u −1
Example of Substitution
EX4: Find ∫ x e dx
2 x3
2
3x
let u =x , du = dx
3
1 x3
1 u
2
x e dx = ∫ e (3 x dx) = ∫ e du
∫
Æ
3
3
2 x3
1 u
1 x3
= e +C = e +C
3
3
Example of Substitution
EX5: Find
∫
( 2 x − 3 ) dx
x2 − 3x
let u = x 2 − 3 x , du = ( 2 x − 3) dx
1
( 2 x − 3) dx
Æ ∫ 2
x − 3x
= ∫ u −1du = ln u + C = ln x 2 − 3 x + C
EX6 Substitution
Find
∫x
1 − xdx
Let u = 1 − x , du = −dx , x=1-u
∫x
(
)
1 − xdx = ∫ (1 − u )u ( − du ) = − ∫ u − u du
1
2
=−
u
1
2
1 +1
2
+1
+
u
3
2
3 +1
2
1
2
3
2
2 32 2 52
+C = − u + u +C
3
5
+1
EX Integrals
Find
4
30
3
(
x
+
3
x
)
(4
x
+ 3)dx
∫
Let g ( x ) = x + 3 x
4
3
′
g ( x) = 4 x + 3
4
30
3
30
x
x
x
dx
g
(
+
3
)
(
4
+
3
)
=
∫
∫ ( x) g ′( x)dx
=
[ g ( x )]31
31
+C =
( x 4 + 3 x )31
31
+C
Example: Find
3
5
2
(
x
+
6
x
)
(
6
x
+ 12)dx
∫
Let u = x + 6 x
3
du = (3 x + 6) dx
2
(6 x 2 + 12) dx = 2(3 x 2 + 6) dx = 2du
3
5
2
5
(
x
+
6
x
)
(
6
x
+
12
)
dx
=
u
∫
∫ 2du
( )+ C
=2
u6
6
=2
( x3 +6 x )6
6
(
)+ C
EX Integrals
Find
∫ sin
Let
g(x)=sin x
10
x cos xdx
g ′( x ) = cos x
10
10
sin
x
cos
xdx
=
g
∫
∫ ( x ) g ′( x )dx
sin11 x
g 11 ( x )
+C
=
+C =
11
11
2
10
(
x
+
4
)
xdx
∫
Example: Find
du = 2 xdx
u = x +4
2
∫ (x
2
+ 4) xdx = ∫ ( x + 4) ⋅ ⋅ 2xdx
10
2
=
1
2
∫u
10 1
2
du =
10
1 u11
2 11
+C
Example: Find ∫ ( x + 4) xdx
2
du = 2 xdx
u = x +4
2
∫ (x
2
10
+ 4) xdx = ∫ ( x + 4) ⋅ ⋅ 2xdx
10
2
=
1
2
=
10 1
2
u
du
=
∫
10
( x2 +4)11
22
+C
1 u11
2 11
+C
Example: Find
2 2
(
+
3
)
x dx
∫
x2
2
It’s not suitable to apply the
substitution rules
2
2
(
+
3
)
x
dx
=
(
+
3
x
+
9
)
x
dx
∫
∫
x2
2
2 2
x4
4
x6
= ∫ ( + 3 x 4 + 9 x 2 )dx
4
x7 3 5
=
+ x + 3x 3 + C
28 5
Substitution Method
The choice of u is one of the following:
1. The quantity under a root or raised to a
power
2. The exponent on e
3. The quantity in the denominator