chapter 5 integrals

Areas and DistancesThe Distance Problem: find the distance traveled by an objectduring a certain time period if the velocity of the object varies.Problem: Suppose the odometer on our car

5 INTEGRALS

5.1 Areas and Distances

5.2 The Definite Integral

Discovery Project: Area Functions

5.3 The Fundamental Theorem of Calculus

5.4 Indefinite Integrals and the Net Change Theorem

(Tích phân bất định và định lý biến thiên toàn phần)

Writing Project: Newton, Leibniz, and the Invention of Calculus

5.1 Areas and Distances

SUMS AND SIGMA NOTATION

We use the symbol Σ to represent a sum, for example

,

2 1

1

2 1

i

n n

i

i

a a

a a

a a

a a

5.1 Areas and DistancesEvaluating Sums

1

, 4

) 1

( )

4 (

, 6

) 1 2

)(

1

(

3 2

1 )

3 (

, 2

) 1

(

3 2 1 )

2 (

, 1

) 1 (

2 2

1 3

2 2

2 1

2 1 1

n n

n n

i

n

n n

i n

n n

i

5.1 Areas and Distances

The Basic Area Problem: find the area of a region S.

S

y = f(x)

Problem: Find the area of a region S lying under the graph y = f(x) ≥ 0

and continuous, above the x – axis and between the vertical lines x = a

5.1 Areas and Distances

PROCEDURE

(1) Divide [a, b] into n equal subintervals by using division points:

a = x0 < x1 < x2 < … < x n – 1 < x n = b the width of each subinterval is Δx = (b – a)/n, x i = a + iΔx.

S1 S2

S

S i

5.1 Areas and Distances

(2) Build rectangles with base length Δx and height f(x i ), i = 1, 2, …, n The area of each rectangle is f(x i )Δx.

a x1 x2 x i – 1 x i x n – 1 b

f(x i)

Δx

5.1 Areas and Distances(3) Add area of all rectangles

5.1 Areas and Distances

5.1 Areas and Distances

Remark!

(1) We get the same value if we use left endpoints

(2) A more general expression for the area is

with (sample point) is any point in the i-th subinterval [x i – 1 , x i]

] )

(

) ( )

( [ lim

(

) ( )

( [ lim

5.1 Areas and Distances

Ex. Find the area under the parabola y = x2, above the x-axis and from x = 0 to x = 1.

Sol. Δx = 1/n, x i = 0 + iΔx = i/n,

n n

n n

21

(

n n

)12

)(

1(







 n n n

5.1 Areas and Distances

The Distance Problem: find the distance traveled by an objectduring a certain time period if the velocity of the object varies

Problem: Suppose the odometer on our car is broken and we want toestimate the distance driven over a 30-second time interval We takespeedometer readings every five seconds and record them in thefollowing table

Time(s) 0 5 10 15 20 25 30

5.1 Areas and Distances

Solution. During each time interval we can estimate the distance byassuming that the velocity is constant

By taking the velocity at the starting point of each time interval, weobtain an estimate for the total distance travelled:

(25×5) + (31×5) + (35×5) + (43×5) + (47×5) + (46×5) = 1135 ft

5.1 Areas and Distances

By taking the velocity at the end point of each time interval, we obtain

an estimate for the total distance travelled:

(31×5) + (35×5) + (43×5) + (47×5) + (46×5) + (41×5) = 1135 ft

For more accurate estimation, we need to take velocity readingsevery two seconds, or every second, or every half of second, etc

5.1 Areas and Distances

The area of each rectangle can be interpreted as a distancebecause the height represents velocity and the width represents time

25

5 10 15 20 25 30 t

v Velocity function

5.1 Areas and Distances

In general, suppose that an object moves with velocity v = f(t), where

a ≤ t ≤ b and f(t) ≥ 0 We take velocity readings at times t0 = a, t1, …, t n =

b so that the velocity is approximately constant on each subinterval If

these times are equally spaced, then the time between consecutive

readings is Δt = (b – a)/n.

The exact distance d traveled is the following limit

t t

f t

t f d

n

i

i n

( lim

5.1 Areas and Distances

Conclusion:

(1) The distance traveled is equal to the area under the graph of the

velocity function

(2) Some other quantities of interest in the natural and social

sciences—such as the work done by a variable force or the cardiacoutput of the heart—can also be interpreted as the area under acurve

5.2 The Definite Integral

Def. Let f(x) be a function defined on [a, b] We divide [a, b] into n equal subintervals of width Δx = (b – a)/n by the division points a = x0, x1,

…x n – 1 , x n = b Let be sample point in each subinterval [x i – 1 , x i] Thenthe definite integral of f from a to b is

provided that this limit exists and gives the same value for all possible

choices of sample points If it does exist, we say that f is integrable

on [a, b].

x x

f dx

x f I

n

i

i n

) (

*

i

x

5.2 The Definite Integral

Note 1. The symbol is called an integral sign, f(x) is called the

integrand, a – lower limit, b – upper limit of integral Theprocedure of calculating an integral is called the integration, dx – indicate the independent variable is x.

Note 2. The definite integral is a number and does not depend on x.

We could use any letter in the place of x.



) ( )

( )

t f dx

x f

5.2 The Definite Integral

Note 3. The sum

is called a Riemann sum

Note 4. If f takes on both positive and negative values then

x x

2 1

) ( x dx A A f

5.2 The Definite Integral

Note 5. There are situations in which it is advantageous to workwith subintervals of unequal width In these cases we have to ensurethat all subinterval widths approach 0 in the limiting process In thiscase the definition of definite integral becomes

Note 6. If f has only a finite number of jump discontinuities on [a, b], then is integrable on [a, b].

x b

a

x x

f dx

x f

I

* 0

) (

5.2 The Definite Integral

Theorem. If f is integrable on [a, b] then  

b

a

x x

f dx

x

f

1

) ( lim

) (

n

a

b i a x i a

x n

a

b

x   i      

5.2 The Definite Integral

Ex1. Express the limit  as a definite integral

2 lim

(2i – 1)/n 1/n

1

2 1

2 lim

2

0

3 1

n

i n

5.2 The Definite Integral

Ex2. Evaluate the Riemann sum for f(x) = x3 – 6x, taking the sample points to be right endpoints and a = 0, b = 3, n = 6.

2

0

, 2

1 6

0

x i x

1

3 6

1

3

1

) 24

( 16

1 8

24 2

1 2

1 2

6 2

)

(

i i

9375

3 2

) 1 6 (

6 24 4

) 1 6 ( 6 16

5.2 The Definite IntegralTHE MIDPOINT RULE

)]

(

) ( )

( [ )

( )

1

n n

i

i b

a

x f x

f x

f x x

x f dx

5.2 The Definite Integral

Ex3. Use the midpoint rule with n = 5 to approximate dx

x

21

1

10

9 2

5

1 2

1

2 1

, 5

1 5

0 19

10 17

10 15

10 13

10 11

10 5

1 1

5.2 The Definite Integral

Properties of the definite integral

5.2 The Definite Integral

function even

) ( ,

) ( 2

) ( )

9

(

function odd

) ( ,

0 )

( )

8

(

, )

( )

( )

7

(

, ) ( )

( then

) ( )

( If )

6

(

, ) ( )

( )

( )

5

(

0

x f dx

x f dx

x f

x f dx

x f

dx x

f dx

x f

dx x g dx

x f x

g x

f

dx x f dx

x f dx

x f

a a

5.3 The Fundamental Theorem of Cal.

The Fundamental Theorem of Calculus establishes a connectionbetween the two branches of calculus: differential calculus and integralcalculus which are inverse processes

The Fundamental Theorem enabled them to compute areas andintegrals very easily without having to compute them as limits of sums

5.3 The Fundamental Theorem of Cal.

PART I.

Consider a function defined by an equation of the form

where f is continuous on [a, b] and x varies between a and b If x is a

fixed number, then the integral is a definite number If x varies,

the number also varies and defines a function of x, denoted by g(x).

) ( )

(  x

a

dt t f x

g

x

a

dt t

f ) (

x

a

dt t

f ) (

5.3 The Fundamental Theorem of Cal.

If f is a positive function, then g(x) can be interpreted as the area under the graph of f from a to x, where x can vary from a to b.

Area = g(x)

y = f(t)

t

5.3 The Fundamental Theorem of Cal.

Ex. Let

(a) Find the values of g(0), g(1), g(2), g(3), and g(4).

(b) Sketch a rough graph of g.

(  x

a

dt t f x

5.3 The Fundamental Theorem of Cal.

(b) As f(t) > 0 for t < 3 we keep adding area for t < 3 and so g is increasing up to x = 3 where it attains a maximum value 4.3 For x > 3,

g decreases because f(t) is negative.

x g

5.3 The Fundamental Theorem of Cal.

) ( )

( lim )

(

5.3 The Fundamental Theorem of Cal.

The Fundamental Theorem of Calculus, Part 1

If f is continuous on [a, b], then the function g defined by

where a ≤ x ≤ b is continuous on [a, b] and differentiable on (a, b), and



 x

a

dt t f x

g ( ) ( )

).

( )

( )

(

dx

d x

5.3 The Fundamental Theorem of Cal.

Ex1. Find the derivative of function

 x t dt x

g

0

2

1 )

(

5.3 The Fundamental Theorem of Cal.

Ex2. Consider the Fresnel function appeared in Fresnel’s theory ofdiffraction of light waves

5.3 The Fundamental Theorem of Cal.

Ex3. Find the derivative of following functions

, )

( )

b (

, )

( )

a (

5

4 2

t

dt e

x x

G

dt e

x F

5.3 The Fundamental Theorem of Cal.

We have the following two additional formulas

).

( ' )) ( ( )

( ' )) ( ( )

( )

2 (

).

( ' )) ( ( )

( )

1 (

) (

) (

) (

x h x h f x

g x g f dt

t

f dx

d

x g x g f dt

t

f dx

d

x g

x h

x g

5.3 The Fundamental Theorem of Cal.

Examples

(1) Find the derivative of

(2) Solve the integral equation:

x x

t

dt e x

H

) ( 3

2 )

f

5.3 The Fundamental Theorem of Cal.

PART II.

The Fundamental Theorem of Calculus, Part 2

If F(x) is any antiderivative of f(x) on [a, b], then we have

).

( )

( )

( )

( x dx F x F b F a f

5.3 The Fundamental Theorem of Cal.

In the physical terms:

If v(t) is the velocity of an object and s(t) is its position at time t, then s’(t) = v(t) Suppose that the object always move in the positive

direction then the area under the velocity curve is equal to the distancetravelled

) ( )

( )

5.3 The Fundamental Theorem of Cal.

Ex1 Evaluate

Ex2. Is the result exact?

Ex3. Evaluate

.)23

2 cos

1 lim



5.3 The Fundamental Theorem of Cal.

From then two fundamental theorems of calculus, it is obviously thatdifferential calculus and integral calculus are inverse processes Eachundoes what the other does

).

( )

( )

( '

), ( )

(

a F b

F dx

x F

x f dt

t

f dx d

5.4 The Indefinite Int & the Net Change Theo.

Indefinite integrals (for more information see 4.8)

The notation is traditionally used for an antiderivative of f

and is called an indefinite integral, hence

For example, we can write

 f ( x ) dx

) ( )

( ' )

( )

3

2

3 3

2

x C

x C

x dx

5.4 The Indefinite Int & the Net Change Theo.

APPLICATION

The fundamental theorem 2 says that if f is continuous on [a, b] then

where F(x) is any antiderivative of f This means F ’ = f, so the above

equation can be rewritten as

) ( )

( )

( )

5.4 The Indefinite Int & the Net Change Theo.

We know that F ’(x) represents the rate of change of y = F(x) with respect to x and F(b) – F(a) is the change in y when x changes from a

to b As y can change in both directions (increase or decrease), F(b) – F(a) represents the net change in y Thus the FTC 2 is

Net Change Theorem: The integral of a rate of change is the netchange

) ( )

( )

5.4 The Indefinite Int & the Net Change Theo.

Some applications

(1) If V(t) is the volume of water in a reservoir at time t, then its

derivative V ’(t) is the rate at which water flows into the reservoir at time t So

is the change in the amount of water in the reservoir between time t1and time t2

) ( )

( )

V dt

t V

5.4 The Indefinite Int & the Net Change Theo.

(2) If [C](t) is the concentration of the product of a chemical reaction

at time t, then the rate of reaction is the derivative d[C]/dt So

is the change in the concentration of C from time t1 to time t2

) ](

[ ) ](

[

]

[

1 2

2

1

t C t

C

dt dt

C d

5.4 The Indefinite Int & the Net Change Theo.

(3) If the mass of a rod measured from the left end to a point x is m(x), then the linear density is ρ(x) = m’(x) So

is the mass of the segment of the rod that lies between x = a and x = b.

) ( )

( )

5.4 The Indefinite Int & the Net Change Theo.

(4) If an object moves along a straight line with position function s(t), then its velocity is v(t) = s’(t), so

is the net change of position, or displacement, of the particle during

the time period from t1 to t2

) ( )

( )

2

1

t s t

s dt

t v

5.4 The Indefinite Int & the Net Change Theo.

A1

A2

A3v

t

3 2

|

, )

(

A A

A dt

t v

A A

A dt

t v

5.4 The Indefinite Int & the Net Change Theo.

Ex1 A particle moves along a line so that its velocity at time t is

v(t) = t2 – t – 6 (measured in meters per second).

(a) Find the displacement of the particle during the time period [1, 4].(b) Find the distance traveled during this time period

5.4 The Indefinite Int & the Net Change Theo.

Solution (a)

This means that the particle moved 4.5 m toward the left

(b) v(t) = t2 – t – 6 = (t – 3)(t + 2) The distance travelled is

2

9 )

6 (

10 6

61 )

6 (

) 6 (

| 6

|

4

3

2 3

1

2 4

1

5.4 The Indefinite Int & the Net Change Theo.

Ex2 Figure shows the power consumption in a certain city for a day(P is measured in megawatts; t is measured in hours starting at midnight) Estimate the energy used on that day

5.4 The Indefinite Int & the Net Change Theo.

Solution. Power is the rate of change of energy P(t) = E ’(t) So by

the Net Change Theorem the total amount of energy used on that dayis

Use the Midpoint Rule to estimate the total energy used

) 0 ( )

24 ( )

(

24 0

E E

dt t



5.5 The Substitution Rule

The method of substitution is the integral version of the chain rule and

Let u = g(x) then du/dx = g’(x), or in differential form, du = g’(x) dx

Thus

))

( ( )

( ' )) ( (



))

( ( )

( )

( ' )

( ' )) ( (



5.5 The Substitution Rule

Substitution in a definite integral

Suppose that g is a differentiable function on [a, b] that satisfies g(a)

= A and g(b) = B Also suppose that f is continuous on the range of g

Then

) ( )

( ' )) (

f dx

x g x g f

5.5 The Substitution Rule

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