Chapter 5 Integration(积分)
In this Chapter, we will encounter some important
concepts
 Antidifferentiation: The Indefinite Integral(不定
积分)
 Integration by Substitution(换元积分法)
 The Definite Integral(定积分) and the
Fundamental Theorem of Calculus(微积分基本定
理)
Section 5.1 Antidifferentiation: The
Indefinite Integral(不定积分)
Antidifferentiation: A function F(x) is said to be an antiderivative of
f(x) if
F ' ( x)  f ( x)
for every x in the domain of f(x). The process of finding
antiderivatives is called antidifferentiation or indefinite integration.
The Constant Rule
The logarithmic rule
The exponential rule
1 kx
 e dx  k e  C
kx
for constant k  0
Note: Notice that the logarithm rule “fills the gap” in the
power rule; namely, the case where n=-1. You may wish to
blend the two rules into this combined form:
 x n 1

 C if n  1
n
 x dx   n  1

ln x  C if n  1
Example 1
Find these integrals:
a.  3dx
b.  x dx
17
c.

1
dx
x
3 x
e
dx
d. 
Solution:
a. Use the constant rule with k=3:
 3dx =3x+C
1
17
18
x
dx

x
C
b. Use the power rule with n=17: 
18
c. Use the power rule with n=-1/2, since n+1=1/2
dx
1 1/ 2
1/ 2
 x   x dx  1 / 2 x  C  2 x  C
d. Use the exponential rule with k=-3:
e
3 x
1
dx 
e 3 x  C
3
Example 2
 x3  2 x  7 

dx
Find the following integral:  


x


Solution:
There is no “quotient rule” for integration, but at least in this case,
you can still divide the denominator into the numerator and then
integrate using the power rule in conjunction with the sum rule,
constant rule and logarithmic rule.
 x3  2 x  7 
7
 2
  x dx    x  2  x dx
1 3
 x  2 x  7 ln x  C
3
Example 3
Solution:
A Differential equation(微分方程) is an equation that involves
differentials or derivatives.
An initial Value problems(初值问题) is a problem that involves
solving a differential equation subject to a specified initial
condition. For instance, we were required to find y=f(x) so that
We solved this initial problem by finding the antiderivative
And using the initial condition to evaluate C.
Example 4
The population p(t) of a bacterial colony t hours after observation
begins is found to be change at the rate
If the population was 2000,000 bacteria when observations
began, what will be population 12 hours later?
Solution:
to be continued
Application of differential equation: Example 5
Newton's law of cooling:
If an object is at a
temperature T at time t and the surrounding is at a constant
temperature Te then the rate of change in T is
where k > 0 is a constant of proportionality that depends properties
of the object.
Change variable to y = T – Te
The equation becomes: dy/dt = -ky
Integrating once we get: y = T – Te = C exp(-kt)
Apply initial condition: T0 – Te = C where T0 is the
initial temperature.
Solution: T(t) = Te + (T0 – Te) exp(-kt)
Exercise 5
Suppose that an ingot leaves the forge at a temperature of 600
Celsius to a room temperature 40 Celsius. It cools to 300
Celsius in 1 hour. How many hours does it take to cool from to
100 degree?
Solving differential equation of the form:
dy/dt = -ky
with k a constant
dy/y = -k dt
Integrating once: ln(y) = -k t + c
Therefor y = exp ( - k t +c ) = C exp ( - k t)
Example 6
Solving the first order differential equation:
dy/dt = -ky3
with k a constant
Example 7
Second order differential equation:
d2y/dt2 = -g
or
with g a constant
y’’ = - g
Integrating once y’ = -gt + c1
y = -gt2/2 + c t + c
The solution is
1
2
where c1 & c2 are integrating constants determined by
initial constants
Remember the gravity acceleration, free falling bodies
problems!
Example 8
Solving the second order differential equation:
d2y/dt2 = -ky
or
with k a constant
y’’ = - ky
y = a sin( bt + c )
The solution is
where b*b = k, a & c are integrating constants
determined by initial constants
This is the simple harmonic motion. All oscillations are
described by this type of DEs.
Section 5.2 Integration by Substitution
How to do the following integral?
Answer
Let
Think of u=u(x) as a change of variable whose differential is
Then
G is an antiderivative of g
Example 9
Find
Solution:
You make the substitution
to obtain
with
Example 10
Solution:
You make the substitution
to obtain
with
Example 11
Solution:
to be continued
Example 12
Solution:
Example 13
Solution:
Example 14
The price p (dollars) of each unit of a particular commodity is
estimated to be changing at the rate
dp
 135 x

dx
9  x2
where x (hundred) units is the consumer demand (the number of
units purchased at that price). Suppose 400 units (x=4) are
demanded when the price is $30 per unit.
a. Find the demand function p(x)
b. At what price will 300 units be demanded? At what price will
no units be demanded?
c. How many units are demanded at a price of $20 per unit?
Solution:
a. The price per unit demanded p(x) is found by integrating p’(x)
with respect to x. To perform this integration, use the substitution
u  9  x2 ,
to get
p ( x)  
du  2 xdx,
xdx 
1
du
2
 135 x
 135  1 
dx   1/ 2  du
u 2
9  x2
1/ 2

 135 1/ 2
 135 u 

  C

u du 

2
2  1/ 2 
 135 9  x  C
2
substitue 9  x for u
2
to be continued
Since p=30 when x=4, you find that
30  135
9  42  C
C  30  135
25  705
so
p ( x )  135
9  x 2  705
b. When 300 units are demanded, x=3 and the corresponding
price is p(3)  135 9  32  705  $132.24 per unit
No units are demanded when x=0 and the corresponding price is
p (0)  135 9  0  705  $300 per unit
c. To determine the number of units demanded at a unit price of
$20 per unit, you need to solve the equation
 135 9  x 2  705  20  x  4.09
That is, roughly 409 units will be demanded when the price is
$20 per unit.
Section 5.3 The Definite Integral and the
Fundamental Theorem of Calculus
Our goal in this section is to show how area under a
curve can be expressed as a limit of a sum of terms
called a definite integral(定积分).
All rectangles
have same width.
n subintervals:
Subinterval width
Formula for xi:
Choice of n evaluation points
Right-endpoint approximation
left-endpoint approximation
Midpoint Approximation
Example 15
=0.285
to be continued
=0.39
=0.33
Example 16
left-endpoint approximation
to be continued
Right-endpoint approximation
to be continued
Midpoint Approximation
S200  1.098608585
S400  1.098611363
Area Under a Curve Let f(x) be continuous and satisfy f(x)≥0 on
the interval a≤x≤b. Then the region under the curve y=f(x) over
the interval a≤x≤b has area
n
A  lim Sn  lim[ f ( x1 )  f ( x2 )  ...  f ( xn )]x  lim  f ( x j )x
n
n
n
j 1
Where x j is the point chosen from the jth subinterval if the Interval
ba
a≤x≤b is divided into n equal parts, each of length x  n
Riemann sum
Let f(x) be a function that is continuous
on the interval a≤x≤b. Subdivide the interval a≤x≤b into n
ba

x

equal parts, each of width
, and choose a number
n
xk from the kth subinterval for k=1, 2, …, . Form the sum
 f ( x1 )  f ( x2 )        f ( xn )x
Called a Riemann sum(黎曼和).
Note: f(x)≥0 is not required
The Definite Integral
the definite integral of f on the
interval a≤x≤b, denoted by  f(x)dx
, is the limit of
the Riemann sum as n→+∞; that is
b
a
The function f(x) is called the integrand, and the
numbers a and b are called the lower and upper limits
of integration, respectively. The process of finding a
definite integral is called definite integration.
Note: if f(x) is continuous on a≤x≤b, the limit used to define
b
integral a f(x)dx exist and is same regardless of how the
subinterval representatives xk are chosen.
Area as a Definite Integral: If f(x) is continuous
and f(x) ≥0 on the interval a≤x≤b, then the region
R under the curve y=f(x) over the interval
a≤x≤b has area A given by the definite integral
b
A   f ( x)dx
a
If f(x) is continuous and f(x)≥0 for all
x in [a,b],then b

f ( x)dx  0
a
and equals the area of the region
bounded by the graph f and the x-axis
between x=a and x=b
If f(x) is continuous and f(x)≤0 for all x
b
in [a,b],then
b

a
f ( x)dx  0
And  f ( x)dx equals the area of the
a
region bounded by the graph f and the
x-axis between x=a and x=b

b
a
f ( x)dx equals the difference between the area under the graph
of f above the x-axis and the area above the graph of f below the
x-axis between x=a and x=b.
This is the net area of the region bounded by the graph of f and
the x-axis between x=a and x=b.
Example 17
The Fundamental Theorem of Calculus If
the function f(x) is continuous on the interval
a≤x≤b, then

b
a
f ( x)dx  F (b)  F (a)
Where F(x) is any antiderivative of f(x) on
a≤x≤b .
Another notation:

b
a
f ( x)dx  F ( x) |ba  F (b)  F (a)
In the case of f(X)≥0, ab f ( x)dxrepresents the area the
curve y=f(x) over the interval [a,b]. For fixed x between
a and b let A(x) denote the area under y=f(x) over the
interval [a,x].
By the definition of the derivative,
Example 18
Solution:
Example 19
Solution:
Integration Rule
Subdivision
Rule
Subdivision Rule
Example 20
Solution:
Example 21
Solution:
to be continued
to be continued
Substituting in a definite integral
Example 22
Option 1.
Option 2.
x2
1 1
2
2 3
dx  
du 
u
x 1
3
3
3
u
x3  1


2
0
x
2
2
2 3
4
dx 
x 1 
3
3
x3  1
0
+C
Substituting in a definite integral
Example 23
to be continued
Example 24
Solution:
to be continued
Section 5.4 Applying Definite
Integration: Area Between Curves
and Average Value
Area Between Curves
Average value of a Function
A Procedure for using Definite Integration in Applications
Section 5.4: Applying definite
integration.
Section 5.4: Applying definite
integration.
Section 5.4: Applying definite
integration.
Section 5.4: Applying definite
integration.
Top curve
Bottom
curve
Area Between Curves
Section 5.4: Applying definite
integration.
The Area Between Two Curve If f(x) and g(x)
are continuous with f(x)≥g(x) on the interval
a≤x≤ b, then the area A between the curves
y=f(x) and y=g(x) over the interval is given by
Section 5.4: Applying definite
integration.
Example 25
Find the area of the region R enclosed by the curves y  x 3
and y  x 2
Solution:
To find the points where the curves intersect, solve the
equations simultaneously as follows:
x3  x 2
x3  x 2  0
x 2 ( x  1)  0
x  0,1
The corresponding points (0,0) and (1,1) are the only
points of intersection.
to be continued
Section 5.4: Applying definite
integration.
The region R enclosed by the two curves is bounded above by y  x
3
y

x
and below by
, over the interval 0  x  1 (See the Figure).
The area of this region is given by the integral
1 3 1 4 1
A   ( x  x )dx   x  x 
0
4
3
0
1
2
3
1
1
1
1
 1

  (1) 3  (1) 4    (0) 3  (0) 4  
4
4
3
 3
 12
Section 5.4: Applying definite
integration.
2
Example 26
Solution:
Section 5.4: Applying definite
integration.
to be continued
Section 5.4: Applying definite
integration.
Example 27
Solution:
Section 5.4: Applying definite
integration.
to be continued
Section 5.4: Applying definite
integration.
Example 28
Find the area of the region bounded by the graph of y=x2
and y=x+2.
Solution:
Section 5.4: Applying definite
integration.
Example 29
Find the area of the region bounded by the graph
of y  2 x , y=0, and y  3 x  5 .
Solution:
Section 5.4: Applying definite
integration.
Average value of a Function(函数的平均值)
Suppose that f is continuous on [a,b] , what is average value of the
function f(x) over the interval a≤x≤b?
1. Subdivide the interval a≤x≤b into n equal parts
2. Choose xj from the jth subinterval for j=1,2…,n. Then the
average of corresponding functional value f(x1), f(x2), …f(xn)
is
Riemann Sum
3. Refine the partition of the interval a≤x≤b by taking
more and more subdivision Points. Then Vn becomes
more and more like the average value of V over the
interval [a,b].The average value V can be think as the
limit of the Riemann sum Vn as n→∞. That is ,
The average value of a Function Let f(x) be a
function that is continuous on the interval a≤x≤b.
Then the average value V of f(x) over a≤x≤b is
given by the definite integral
Example 30
Solution:
Geometric Interpretation of Average Value The average value V
of f(x) over an interval a≤x≤b where f(x) is continuous and
satisfies f(x)≥0 is equal to the height of a rectangle whose base is
the interval and whose area is the same as the area under the
curve y=f(x) over a≤x≤b .
The rectangle with base a≤x≤b and
height V has the same area as the
region under the curve y=f(x) over
a≤x≤b .
Section 5.5 Additional Applications to
Business and Economics
Future Value and Present Value of an Income Flow
 Term: A specified time period 0≤t≤T.
 An Income Flow (stream)(现金流): A stream of income
transferred continuously into an account.
 Future value of the income stream: Total amount (money
transferred into the account plus interest) that is accumulated
During the specified term
 Annuity(年金): The strategy is to approximate the
continuous income stream by a sequence of discrete
deposits.
Example 31
Money is transferred continuously into an account at the constant
rate of $1200 per year. The account earns interest at the annual rate
of 8% compounded continuously. How much will be in the account
at the end of 2 years?
Recall:
Step 1. Thus P dollars invested at 8% compounded
continuously will be worth Pe0.08t dollars t years later
Section 5.5 Additional
Applications to Business.
to be continued
Step 2. Divide the 2-year time interval 0≤t≤2 into n equal
Subinterval of length ∆t years and let tj denote the beginning
of the jth subinterval. Then, during the jth subinterval
Money deposited=(dollars per year) (Number of years)=1200∆t
Step 3. If all of this money were deposited at the beginning of
the subinterval, it would remain in the account for 2-tj years and
therefore would grow to (1,200t )e 0.08( 2t ) dollars. Thus,
j
Future value of money deposited
during jth subinterval
≈1200e0.08(2-tj)∆t
Section 5.5 Additional
Applications to Business.
to be continued
Step 4. The future value of the entire income stream is the
sum of the future values of the money deposited during each
of the n subintervals. Hence
Step 5. As n increase without bound, the length of each
subinterval approaches zero and the approximation
approaches the true future value of the income stream. Hence
Section 5.5 Additional
Applications to Business.
Future Value of an Income Stream Suppose money is
being transferred continuously into an account over a time
period 0≤t≤T at a rate given by the function f(t) and that the
account earns interest at an annual rate r compounded
continuously. Then the future value of FV of the income
stream over the term T is given by the definite integral
Section 5.5 Additional
Applications to Business.
Present value of an income stream: The amount of money
A that must be deposited now at the prevailing interest rate to
generate the same income as the income stream over the
same T-year period.
Generated at a continuous rate f(x)
Since A dollars invested
at annual interest rate r
compounded continuously
will be worth Aert dollars
in T years
Section 5.5 Additional
Applications to Business.
Present Value of an Income Stream The
present value PV of an income steam that is
deposited continuously at the rate f(t) into
an account that earns interest at an annual
rate r compounded continuously for a term
of T years is given by
Section 5.5 Additional
Applications to Business.
Example 32
Jane is trying to decide between two investments. The first
costs $1000 and is expected to generate a continuous
income stream at the rate of f1(t)=3000e0.03t dollars per
year. The second investment costs $4000 and is estimated
to generate income at the constant rate of f2(t)=4000
dollars per year. If the prevailing annual interest rate
remains fixed at 5% compounded continuously over the
next 5 years, which investment will generate more net
income over this time period?
Section 5.5 Additional
Applications to Business.
Solution:
Section 5.5 Additional
Applications to Business.
to be continued
Section 5.5 Additional
Applications to Business.
Section 5.6 Additional Applications to
the Life and Social Sciences
The Volume(体积) of a Solid of Revolution
The volume of a Solid of Revolution
Section 5.6 Additional
Applications.
Cross-sections perpendicular to the
axis of rotation are circular
So a typical slice may be viewed as a
thin disk
Solids with cross-sectional area A(x)
A   r 2   [ f ( x )]2
Divide the interval a≤x≤b into n equal
subintervals of length △x
Section 5.6 Additional
Applications.
The total volume V can be approximated by
n
2
   [ f ( x )] x
i
i 1
The approximation improves as n increase without bound (△x
n
approach 0) and
V  lim   [ f ( xi )]2 x
n 
i 1
=  [ f ( x )] dx   A( x )dx
b
a
2
b
a
Section 5.6 Additional
Applications.
Volume Formula
Suppose f(x) is continuous and f(x) ≥0 and a≤x≤b and let R be
the region under the curve y=f(x) between x=a and x=b. Then
the solid S formed by revolving R about the x axis has volume
b
Volume of S    [ f ( x)]2 dx
a
Section 5.6 Additional
Applications.
Example 33
Find the volume of the solid S formed by revolving the region
under the curve y  x 2  1 from x=0 to x=2 about the x axis.
Section 5.6 Additional
Applications.
Solution:
The region, the solid of revolution, and the jth disk are shown in
the Figure. The radius of the jth disk is f ( x j )  x 2j  1 . Hence,
Volume of jth disk   [ f ( x j )]2 x   ( x 2j  1) 2 x
and
Volume of S  lim
n 
n
 (x
j 1
2
j
 1) 2 x
2
   ( x 2  1) 2 dx
0
2
   ( x 4  2 x 2  1) dx
0
1 5 2 3
 2 206
   x  x  x 
  43.14
3
15
5
0
Section 5.6 Additional
Applications.
Chapter 6 Additional Topics in
Integration
In this Chapter, we only talk about Section 6.1.
 Integration by parts(分部积分法).
Section 6.1 Integration by Parts.
If u(x) and v(x) are both differentiable functions of x, then
Since u(x)v(x) is antiderivative of
and
Since
We have
Integration by parts formula:
 f ( x)dx   udv  uv   vdu
Integration by parts:
Step 1. Choose functions u and v so that f(x)dx=udv. Try to
pick u so that du is simpler than u and a dv is easy to integrate
Step 2. Organize the computation of du and v as
and substitute into the integration by parts formula
Step 3. Complete the integration by finding
Then
Example 1
Solution:
to be continued
Example 2
Solution:
Example 3
Find the area of the region bounded by the curve y=lnx, the x axis,
and lines x=1 and x=e.
Solution:
Example 4
Solution:
to be continued
Repeated Application of Integration by parts
Example 5
Solution:
to be continued
Taylor's formula with integral remainder,
derived by Brook Taylor (1685-1731)
By repeating this integration by parts process on the remaining integral p
times, one has the result:
The last term is referred to as the remainder, Rn(x)
118
Application of Taylor series
Evaluate the definite integral :
119
Application of Euler's formula : consider the integral
120