Chapter 5 Integration( 积分 ) In this Chapter, we will encounter some important concepts Antidifferentiation: The Indefinite Integral( 不定积分 ) Integration

Chapter 5 Integration(Chapter 5 Integration( 积分积分 ))

In this Chapter, we will In this Chapter, we will encounter some important encounter some important conceptsconcepts

Antidifferentiation: The Indefinite Integral(Antidifferentiation: The Indefinite Integral( 不定不定积分积分 ))

Integration by Substitution(Integration by Substitution( 换元积分法换元积分法 ))

The Definite Integral(The Definite Integral( 定积分定积分 ) and the ) and the Fundamental Theorem of Calculus(Fundamental Theorem of Calculus( 微积分基本定微积分基本定理理 ) )

Section 5.1 Antidifferentiation: The Indefinite Integral(不定积分 )

AntidifferentiationAntidifferentiation: A function : A function F(x)F(x) is said to be an antiderivative is said to be an antiderivative of of f(x)f(x) if if

for every for every xx in the domain of in the domain of f(x).f(x). The process of finding The process of finding antiderivatives is called antidifferentiation or indefinite integration. antiderivatives is called antidifferentiation or indefinite integration.

)()(' xfxF

The Constant RuleThe Constant Rule

0constant for 1

kCek

dxe kxkx

The logarithmic ruleThe logarithmic rule

The exponential ruleThe exponential rule

NoteNote: : Notice that the logarithm rule “fills the gap” in the Notice that the logarithm rule “fills the gap” in the power rule; namely, the case where power rule; namely, the case where n=-1n=-1. You may wish to . You may wish to blend the two rules into this combined form: blend the two rules into this combined form:

1 if ln

1 if 1

1

nCx

nCn

xdxx

n

n

Example 1

Find these integrals:Find these integrals:

a. b. c. d. a. b. c. d.

dx3 dxx17 dxx

1 dxe x3

Solution:Solution:

a.a. Use the constant rule with Use the constant rule with k=3k=3: =: =3x+C3x+C

b.b. Use the power rule with Use the power rule with n=17n=17::

c.c. Use the power rule with Use the power rule with n=-1/2n=-1/2, since , since n+1=1/2n+1=1/2

d.d. Use the exponential rule with Use the exponential rule with k=-3: k=-3:

dx3

Cxdxx 1817

18

1

CxCxdxxx

dx2

2/1

1 2/12/1

Cedxe xx 33

3

1

Example 2

Find the following integral:Find the following integral:

dxx

xx

723

Solution:Solution:

There is no “quotient rule” for integration, but at least in this case, There is no “quotient rule” for integration, but at least in this case, you can still divide the denominator into the numerator and then you can still divide the denominator into the numerator and then integrate using the power rule in conjunction with the sum rule, integrate using the power rule in conjunction with the sum rule, constant rule and logarithmic rule. constant rule and logarithmic rule.

Cxxx

dxx

xdxx

xx

ln723

1

72

72

3

23

Example 3

Solution:Solution:

A A Differential equation(Differential equation( 微分方程微分方程 )) is an equation that is an equation that involves differentials or derivatives.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 We solved this initial problem by finding the antiderivative

And using the initial condition to evaluate C.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 If the population was 2000,000 bacteria when observations began, what will be population 12 hours later? began, what will be population 12 hours later?

Solution:Solution:

to be continued

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.

Application of differential equation: Example 5

Change variable to y = T – Te

The equation becomes: dy/dt = -kyIntegrating 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)

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?

Exercise 5

dy/y = -k dtIntegrating once: ln(y) = -k t + cTherefor y = exp ( - k t +c ) = C exp ( - k t)

Solving differential equation of the form:

dy/dt = -ky with k a constant

Example 6

Solving the first order differential equation:

dy/dt = -ky3 with k a constant

Example 7

Second order differential equation:

d2y/dt2 = -g with g a constant

or y’’ = - g

Integrating once y’ = -gt + c1

The solution is y = -gt2/2 + c1t + c2where 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 with k a constant

or y’’ = - ky

The solution is y = a sin( bt + c )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 SubstitutionSection 5.2 Integration by Substitution

How to do the following integral?How to do the following integral?

AnswerAnswer

LetLet

Think of Think of uu==uu((xx) as a change of variable whose differential is) as a change of variable whose differential is

ThenThen

GG is an antiderivative of is an antiderivative of gg

Example 9

FindFind

Solution:Solution:

You make the substitution You make the substitution withwith

to obtainto obtain

Example 10

Solution:Solution:

You make the substitution You make the substitution withwith

to obtainto obtain

Example 11

Solution:Solution:

to be continued

Example 12

Solution:Solution:

Example 13

Solution:Solution:

Example 14

The price The price pp (dollars) of each unit of a particular commodity is (dollars) of each unit of a particular commodity is estimated to be changing at the rate estimated to be changing at the rate

29

135

x

x

dx

dp

where where xx (hundred) units is the consumer demand (the number of (hundred) units is the consumer demand (the number of units purchased at that price). Suppose units purchased at that price). Suppose 400400 units ( units (x=4x=4) are ) are demanded when the price is demanded when the price is $30$30 per unit. per unit.

a. Find the demand function p(x)

b.b. At what price will At what price will 300300 units be demanded? At what price will units be demanded? At what price will no units be demanded?no units be demanded?

c.c. How many units are demanded at a price of How many units are demanded at a price of $20$20 per unit? per unit?

Solution:Solution:

a. The price per unit demanded a. The price per unit demanded p(x)p(x) is found by integrating is found by integrating p’(x)p’(x) with respect to with respect to xx. To perform this integration, use the substitution . To perform this integration, use the substitution

duxdxxdxduxu2

1 ,2 ,9 2

to getto get

uforxsubstitueCx

Cu

duu

duu

dxx

xxp

9 9135

2/12

135

2

135

2

1135

9

135)(

22

2/12/1

2/12

to be continued

Since Since p=30p=30 when when x=4x=4, you find that , you find that

7059135)(

so

7052513530

4913530

2

2

xxp

C

C

b. When b. When 300300 units are demanded, units are demanded, x=3x=3 and the corresponding and the corresponding price isprice is

No units are demanded when x=0 and the corresponding price is No units are demanded when x=0 and the corresponding price is

unitper 24.132$70539135)3( 2 p

unitper 300$70509135)0( p

c. To determine the number of units demanded at a unit price of c. To determine the number of units demanded at a unit price of $20$20 per unit, you need to solve the equation per unit, you need to solve the equation

09.4207059135 2 xx

That is, roughly That is, roughly 409409 units will be demanded when the price is units will be demanded when the price is $20$20 per unit. per unit.

Section 5.3 The Definite Integral and the Section 5.3 The Definite Integral and the Fundamental Theorem of CalculusFundamental Theorem of Calculus

Our goal in this section is to show how area under a Our goal in this section is to show how area under a curve can be expressed as a limit of a sum of terms curve can be expressed as a limit of a sum of terms called a called a definite integral(definite integral( 定积分定积分 ))..

All rectangles All rectangles have same widthhave same width.

nn subintervals subintervals:

Subinterval widthSubinterval width

Formula for xFormula for xii::

Choice of Choice of nn evaluation points evaluation points

Right-endpoint approximationRight-endpoint approximation left-endpoint approximationleft-endpoint approximation

Midpoint ApproximationMidpoint Approximation

=0.285=0.285

Example 15

to be continued

=0.39=0.39 =0.33=0.33

Example 16

left-endpoint approximationleft-endpoint approximation

to be continued

Right-endpoint approximationRight-endpoint approximation

to be continued

Midpoint ApproximationMidpoint Approximation

31.09861136 51.09860858 400200 SS

Area Under a CurveArea Under a Curve Let Let ff((xx) be continuous and satisfy ) be continuous and satisfy ff((xx))≥0 on ≥0 on the interval the interval a≤x≤ba≤x≤b. Then the region under the curve . Then the region under the curve y=f(x)y=f(x) over over the interval the interval a≤x≤ba≤x≤b has area has area

1 21

lim lim[ ( ) ( ) ... ( )] lim ( )n

n n jn n n

j

A S f x f x f x x f x x

Where Where is the point chosen from the is the point chosen from the jjth subinterval if the Interval th subinterval if the Interval a≤x≤ba≤x≤b is divided into is divided into nn equal parts, each of length equal parts, each of length

b ax

n

jx

Riemann sumRiemann sum Let Let f(x)f(x) be a function that is continuous be a function that is continuous on the interval on the interval a≤x≤ba≤x≤b. Subdivide the interval . Subdivide the interval a≤x≤ba≤x≤b into into nn equal parts, each of width , and choose a number equal parts, each of width , and choose a number xkxk from the from the kkth subinterval for th subinterval for k=1, 2, …,k=1, 2, …, . Form the sum . Form the sum

b ax

n

Called a Called a Riemann sum(Riemann sum( 黎曼黎曼和和 ).).

xxfxfxf n )()()( 21

Note: Note: f(x)≥0f(x)≥0 is not required is not required

The Definite IntegralThe Definite Integral the definite integral of f on the the definite integral of f on the interval interval a≤x≤ba≤x≤b, denoted by , is the limit of , denoted by , is the limit of the Riemann sum as the Riemann sum as n→+∞n→+∞; that is; that is

b

af(x)dx

The function The function f(x)f(x) is called the is called the integrandintegrand, and the , and the numbers numbers aa and and bb are called the are called the lower and upper limits lower and upper limits of integrationof integration, respectively. The process of finding a , respectively. The process of finding a definite integral is called definite integral is called definite integrationdefinite integration..

Note: if f(x) is continuous on a≤x≤b, the limit used to defineNote: if f(x) is continuous on a≤x≤b, the limit used to defineintegralintegral exist and is same regardless of how the exist and is same regardless of how the subinterval representatives xk are chosen.subinterval representatives xk are chosen.

b

af(x)dx

Area as a Definite IntegralArea as a Definite Integral: If : If f(x)f(x) is continuous is continuous and and f(x) ≥0 f(x) ≥0 on the interval on the interval aa≤x≤b≤x≤b, then the region , then the region RR under the curve under the curve y=f(x)y=f(x) over the interval over the interval

aa≤x≤b ≤x≤b has area has area AA given by the definite integral given by the definite integral

b

adxxfA )(

If f(x) is continuous and f(x)≥0 for all x in [a,b],then

( ) 0b

af x dx

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 in [a,b],then

( ) 0b

af x dx

And equals the area of the region bounded by the graph f and the x-axis between x=a and x=b

( )b

af x dx

equals the difference between the area under the graph equals the difference between the area under the graph

of of f f above the above the xx-axis and the area above the graph of f below the -axis and the area above the graph of f below the x-axis between x-axis between x=a x=a and and x=b.x=b.

( )b

af x dx

This is the net area of the region bounded by the graph of This is the net area of the region bounded by the graph of f f andand the the x-x-axis between axis between x=ax=a and and x=b.x=b.

Example 17

The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus If If the function the function f(x)f(x) is continuous on the interval is continuous on the interval a≤x≤ba≤x≤b, then , then

( ) ( ) ( )b

af x dx F b F a

Where Where F(x)F(x) is any antiderivative of is any antiderivative of f(x)f(x) on on a≤x≤b .a≤x≤b .

Another notation:

( ) ( ) | ( ) ( )b b

aaf x dx F x F b F a

In the case of f(X)≥0, represents 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].

( )b f x dxa

By the definition of the derivative,By the definition of the derivative,

Example 18

Solution:Solution:

Example 19

Solution:Solution:

Integration RuleIntegration Rule

Subdivision Subdivision RuleRule

Subdivision RuleSubdivision Rule

Example 20

Solution:Solution:

Example 21

Solution:Solution:

to be continued

to be continued

Substituting in a definite integralSubstituting in a definite integral

Example 22

Option 1.

Option 2.2

3

3

222 3

300

1 1 2 21

3 3 31

2 41

3 31

xdx du u x

ux

xdx x

x

+C+C

Substituting in a definite integralSubstituting in a definite integral

Example 23

to be continued

Example 24

Solution:Solution:

to be continued

Section 5.4 Applying Definite Integration: Area Between Curves

and Average Value

Area Between CurvesArea Between Curves

Average value of a FunctionAverage value of a Function

Section 5.4: Applying definite integration.

A Procedure for using Definite Integration in Applications




Top curveTop curveBottomBottomcurvecurve


Area Between CurvesArea Between Curves


The Area Between Two CurveThe Area Between Two Curve If If f(x)f(x) and and g(x)g(x) are continuous with are continuous with f(x)≥g(x)f(x)≥g(x) on the interval on the interval a≤x≤ ba≤x≤ b, then the area A between the curves , then the area A between the curves y=f(x)y=f(x) and and y=g(x)y=g(x) over the interval is given by over the interval is given by


Example 25

Find the area of the region Find the area of the region R R enclosed by the curves enclosed by the curves

and and

3xy 2xy

Solution:Solution:

To find the points where the curves intersect, solve the To find the points where the curves intersect, solve the equations simultaneously as follows:equations simultaneously as follows:

1,0

0)1(

02

23

23

x

xx

xx

xx

The corresponding points (0,0) and (1,1) are the only points of intersection.

to be continued


The region R enclosed by the two curves is bounded above by The region R enclosed by the two curves is bounded above by and below by , over the interval (See the Figure). and below by , over the interval (See the Figure). The area of this region is given by the integral The area of this region is given by the integral

2xy 3xy 10 x

12

1)0(

4

1)0(

3

1)1(

4

1)1(

3

1

0

1

4

1

3

1)(

4343

431

0

32

xxdxxxA


Example 26

Solution:Solution:

to be continued



Example 27

Solution:Solution:

to be continued



Find the area of the region bounded by the graph of Find the area of the region bounded by the graph of y=xy=x22 and and y=x+2y=x+2..

Example 28

Solution:Solution:


Example 29

Solution:Solution:

Find the area of the region bounded by the graph of Find the area of the region bounded by the graph of , , y=0y=0, and ., and . 2y x 3 5y x

Average value of a Function(Average value of a Function( 函数的平均函数的平均值值 ))Suppose that Suppose that ff is is continuouscontinuous on on [a,b][a,b] , what is average value of the , what is average value of the

function function f(x)f(x) over the interval over the interval a≤x≤b?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 FunctionThe average value of a Function Let Let f(x)f(x) be a be a function that is continuous on the interval function that is continuous on the interval a≤x≤ba≤x≤b. . Then the average value Then the average value VV of of f(x)f(x) over over a≤x≤ba≤x≤b is is given by the definite integralgiven by the definite integral

Example 30

Solution:Solution:

Geometric Interpretation of Average ValueGeometric Interpretation of Average Value The average value The average value VVof of f(x)f(x) over an interval over an interval aa≤x≤b≤x≤b where where f(x)f(x) is continuous and is continuous and satisfies satisfies f(x)≥0f(x)≥0 is equal to the height of a rectangle whose base is is equal to the height of a rectangle whose base is the interval and whose area is the same as the area under the the interval and whose area is the same as the area under the curve curve y=f(x)y=f(x) over over a≤x≤ba≤x≤b . .

The rectangle with base a≤x≤b and The rectangle with base a≤x≤b and height V has the same area as the height V has the same area as the region under the curve y=f(x) over region under the curve y=f(x) over a≤x≤ba≤x≤b . .

SectionSection 5.5 Additional Applications to 5.5 Additional Applications to Business and EconomicsBusiness and Economics

Future Value and Present Value of an Income FlowFuture 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.

Section 5.5 Additional Applications to Business.

Money is transferred continuously into an account at the constant Money is transferred continuously into an account at the constant rate of rate of $1200$1200 per year. The account earns interest at the annual rate per year. The account earns interest at the annual rate of of 8%8% compounded continuously. How much will be in the account compounded continuously. How much will be in the account at the end of at the end of 22 years? years?

Example 31

Recall:Recall:

Step 1. Thus P dollars invested at 8% compounded continuously will be worth Pe0.08t dollars t years later

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 andtherefore would grow to dollars. Thus,

)2(08.0)200,1( jtet

Future value of money deposited during jth subinterval ≈1200e0.08(2-t

j)∆t

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


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


Present value of an income streamPresent 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


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


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?

Example 32


Solution:Solution:

to be continued


SectionSection 5.6 Additional Applications to 5.6 Additional Applications to the Life and Social Sciencesthe Life and Social Sciences

The Volume(The Volume(体积体积 ) of a Solid of ) of a Solid of RevolutionRevolution

Section 5.6 Additional Applications.

The volume of a Solid of RevolutionThe volume of a Solid of Revolution


Cross-sections perpendicular to the Cross-sections perpendicular to the axis of rotation are circularaxis of rotation are circular

So a typical slice may be viewed as a So a typical slice may be viewed as a thin diskthin disk

Solids with cross-sectional area Solids with cross-sectional area AA((xx))2 2[ ( )]A r f x

Divide the interval a≤x≤b into n equal Divide the interval a≤x≤b into n equal subintervals of length x△subintervals of length x△


The total volume The total volume VV can be approximated by can be approximated by

2[ ( )]1

nf x xi

i

The approximation improves as The approximation improves as nn increase without bound ( increase without bound ( x△x△ approach approach 00) and) and

2

1

2

lim [ ( )]

= [ ( )] ( )

n

ini

b b

a a

V f x x

f x dx A x dx


Volume FormulaVolume Formula

Suppose Suppose f(x)f(x) is continuous and is continuous and f(x) ≥0f(x) ≥0 and and a≤x≤b a≤x≤b and let and let RR be be the region under the curve the region under the curve y=f(x)y=f(x) between between x=ax=a and and x=b. x=b. Then Then the solid the solid SS formed by revolving formed by revolving RR about the about the xx axis has volume axis has volume

b

adxxf 2])([S of Volume


Example 33

Find the volume of the solid Find the volume of the solid SS formed by revolving the region formed by revolving the region under the curve from under the curve from x=0x=0 to to x=2x=2 about the x axis. about the x axis. 12 xy


Solution:Solution:

The region, the solid of revolution, and the The region, the solid of revolution, and the jjth disk are shown in th disk are shown in the Figure. The radius of the the Figure. The radius of the jjth disk is . Hence,th disk is . Hence,1)( 2 jj xxf

xxxxfj jj 222 )1()]([diskth of Volume

andand

14.4315

206

0

2

3

2

5

1

)12(

)1(

)1(lim of Volume

35

2

0

24

2

0

22

1

22

xxx

dxxx

dxx

xxSn

jj

n

Chapter 6 Additional Topics in Chapter 6 Additional Topics in IntegrationIntegration

In this Chapter, we only talk about Section 6.1.In this Chapter, we only talk about Section 6.1.

Integration by parts(Integration by parts( 分部积分法分部积分法 ). ).

Section 6.1 Integration by Parts.If If u(x)u(x) and and v(x)v(x) are both differentiable functions of are both differentiable functions of xx, then, then

Since u(x)v(x) is antiderivative of andSince u(x)v(x) is antiderivative of and

Since Since

We haveWe have

Integration by parts formula:Integration by parts formula:

vduuvudvdxxf )(

Integration by parts:Integration by parts:

Step 1.Step 1. Choose functions Choose functions uu and and v v so that so that ff((xx))dx=udvdx=udv. Try to . Try to pick pick uu so that so that dudu is simpler than is simpler than uu and a and a dvdv is easy to integrate is easy to integrate

Step 2.Step 2. Organize the computation of Organize the computation of dudu and and vv as as

and substitute into the integration by parts formulaand substitute into the integration by parts formula

Step 3Step 3. Complete the integration by finding Then. Complete the integration by finding Then

Example 1

Solution:Solution:

to be continued

Example 2

Solution:Solution:

Find the area of the region bounded by the curve Find the area of the region bounded by the curve y=lnxy=lnx, the , the xx axis, axis,and lines and lines x=1x=1 and and x=ex=e..

Example 3

Solution:Solution:

Example 4

Solution:Solution:

to be continued

Repeated Application of Integration by partsRepeated Application of Integration by parts

Example 5

Solution:Solution:

to be continued

118

By repeating this integration by parts process on the remaining integral p times, one has the result:

Taylor's formula with integral remainder, derived by Brook Taylor (1685-1731)

The last term is referred to as the remainder, Rn(x)

Application of Taylor series Application of Taylor series

119

Evaluate the definite integral :

120

Application of Euler's formula : consider the integral

Documents

Chapter 5 Integration( 积分 ) In this Chapter, we will encounter some important concepts Antidifferentiation: The Indefinite Integral( 不定积 分 ) Integration

Chapter 5 Integration( 积分 ) In this Chapter, we will encounter some important concepts Antidifferentiation: The Indefinite Integral( 不定积分 ) Integration