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 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

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:

Example 1 Find these integrals: a. b. c. d. Solution: Solution: a.Use the constant rule with k=3: =3x+C b.Use the power rule with n=17: c.Use the power rule with n=-1/2, since n+1=1/2 d.Use the exponential rule with k=-3:

Example 2 Find the following integral: Solution: 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.

Example 3 Solution: 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: 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 T e 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 – T e The equation becomes: dy/dt = -ky Integrating once we get: y = T – T e = C exp(-kt) Apply initial condition: T 0 – T e = C where T 0 is the initial temperature. Solution: T(t) = T e + (T 0 – T e ) 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 dt Integrating once: ln(y) = -k t + c Therefor 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 = -ky 3 with k a constant

Example 7 Second order differential equation: d 2 y/dt 2 = -g with g a constant or y’’ = - g Integrating once y’ = -gt + c 1 The solution is y = -gt 2 /2 + c 1 t + c 2 where c 1 & c 2 are integrating constants determined by initial constants Remember the gravity acceleration, free falling bodies problems!

Example 8 Solving the second order differential equation: d 2 y/dt 2 = -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 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: Solution: You make the substitution with to obtain

Example 10 Solution: Solution: You make the substitution with to obtain

Example 11 Solution: Solution: to be continued

Example 12 Solution: Solution:

Example 13 Solution: Solution:

Example 14 The price p (dollars) of each unit of a particular commodity is estimated to be changing at the rate 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: 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 to get to be continued

Since p=30 when x=4, you find that b. When 300 units are demanded, x=3 and the corresponding price is No units are demanded when x=0 and the corresponding price is c. To determine the number of units demanded at a unit price of $20 per unit, you need to solve the equation 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 have same width.  n subintervals  n subintervals :  Subinterval width  Formula for x i :

Choice of n evaluation points

Right-endpoint approximation left-endpoint approximation

Midpoint Approximation

=0.285 Example 15 to be continued

=0.39=0.33

Example 16 left-endpoint approximation to be continued

Right-endpoint approximation to be continued

Midpoint Approximation

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 Where is the point chosen from the jth subinterval if the Interval a≤x≤b is divided into n equal parts, each of length

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 equal parts, each of width, and choose a number xk from the kth subinterval for k=1, 2, …,. Form the sum Called a Riemann sum( 黎曼和 ). 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, is the limit of the Riemann sum as n→+∞; that is 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 integral 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

If f(x) is continuous and f(x)≥0 for all x in [a,b],then 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 And equals the area of the region bounded by the graph f and the x-axis between x=a and x=b

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

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].

By the definition of the derivative,

Example 18 Solution: Solution:

Example 19 Solution: Solution:

Integration Rule SubdivisionRule

Subdivision Rule

Example 20 Solution: Solution:

Example 21 Solution: Solution: to be continued

Substituting in a definite integral

Example 22 Option 1. Option 2. +C

Substituting 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 Curves  Average value of a Function

Section 5.4: Applying definite integration. A Procedure for using Definite Integration in Applications

Section 5.4: Applying definite integration.

Top curve Bottomcurve

Section 5.4: Applying definite integration. 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 and Solution: Solution: To find the points where the curves intersect, solve the equations simultaneously as follows: 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 and below by, over the interval (See the Figure). The area of this region is given by the integral

Section 5.4: Applying definite integration. Example 26 Solution: Solution: to be continued

Section 5.4: Applying definite integration.

Example 27 Solution: Solution: to be continued

Section 5.4: Applying definite integration.

Find the area of the region bounded by the graph of y=x 2 and y=x+2. Example 28 Solution: Solution:

Section 5.4: Applying definite integration. Example 29 Solution: Solution: Find the area of the region bounded by the graph of, y=0, and.

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 x j from the jth subinterval for j=1,2…,n. Then the average of corresponding functional value f(x 1 ), f(x 2 ), …f(x n ) 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 V n 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: 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.

Section 5.5 Additional Applications to Business. 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? Example 31 Recall: Step 1. Thus P dollars invested at 8% compounded continuously will be worth Pe 0.08t dollars t years later to be continued

Section 5.5 Additional Applications to Business. Step 2. Divide the 2-year time interval 0≤t≤2 into n equal Subinterval of length ∆t years and let t j 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-t j years and therefore would grow to dollars. Thus, Future value of money deposited during jth subinterval ≈ 1200e 0.08(2-t j ) ∆t to be continued

Section 5.5 Additional Applications to Business. 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 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 Ae rt 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. 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 f 1 (t)=3000e 0.03t dollars per year. The second investment costs $4000 and is estimated to generate income at the constant rate of f 2 (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

Section 5.5 Additional Applications to Business. Solution: Solution: 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

Section 5.6 Additional Applications. The volume of a Solid of Revolution

Section 5.6 Additional Applications. Cross-sections perpendicular to the axis of rotation are circular axis of rotation are circular So a typical slice may be viewed as a thin disk Solids with cross-sectional area A(x) 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 The approximation improves as n increase without bound ( △ x approach 0) and

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

Section 5.6 Additional Applications. Example 33 Find the volume of the solid S formed by revolving the region under the curve from x=0 to x=2 about the x axis.

Section 5.6 Additional Applications. Solution: Solution: The region, the solid of revolution, and the jth disk are shown in the Figure. The radius of the jth disk is. Hence, and

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:

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: Solution: to be continued

Example 2 Solution: Solution:

Find the area of the region bounded by the curve y=lnx, the x axis, and lines x=1 and x=e. Example 3

Solution: Solution:

Example 4 Solution: Solution: to be continued

Repeated 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 ( ) The last term is referred to as the remainder, R n (x)

Application of Taylor series 119 Evaluate the definite integral :

120 Application of Euler's formula : consider the integral