Chapter 5 Integrals 机动目录上页下页返回结束 5.2 Area 5.3 The Definite Integral 5.4 The Fundamental Theorem of Calculus 5.5 The Substitution Rule

Chapter 5 Integrals

机动目录上页下页返回结束

5.2 Area

5.3 The Definite Integral5.4 The Fundamental Theorem of Calculus

5.5 The Substitution Rule

4.2 AreaArea Problem: Find the area of the region S that lies under the curve y=f(x) from a to b.(see Figure 1)

y=f(x)

S

a b x

y

o

Figure 1


Idea for problem solving: First approximate the region S by polygons, and then take the limit of the area of these polygons.(see the following example)

Example 1 Find the area under the parabola y=x2 from 0 to 1.

ySolution: We start by dividing the interval [0, 1] into n-subintervals with equal length, and consider the rectangles whose bases are these subintervals and whose heights are the values of the function at the right-hand endpoints.

o x

(1,1)

1/n1

Figure 2


Then the sum of the area of these rectangles is

26

)12)(1(21221211 )()()(n

nnnn

nnnnnnS

As n increases, Sn becomes a better and better approximation to the area of the parabolic segment. Therefore we define the area A to be the limit of the sums of the areas of these rectangles, that is,

31lim

nn

SA

Applying the idea of Example 1 to the more general region S of F.1, we introduce the definition of the area as following:

Step 1: Partition--Divide the interval [a, b] into n smaller subintervals by choosing partition points x0 , x1 , x2 ,…., xn so that

a=x0 < x1<x2 <…< xn=b机动目录上页下页返回结束

Step 2: Approximation—By the partition above, the area of S can be approximated by the sum of areas of n rectangles .

This subdivision is called a partition of [a,b] and we denote it by P . Let denote the length of ith subinterval [x

i-1 , xi] , and ||P|| (the norm of P) denotes the length of longest subinterval. Thus

1 iii xxx

},,max{|||| 1 nxxP

Using the partition P one can divide the region S into n strips (see F.3). Now, we choose a number in each subinterval, then each strip Si can be approximated by a rectangle Ri (see F.4).

ix

The sum of areas of these rectangles as an approximation is

n

iii

n

ii xxfA

11

)(


y=f(x)

Si

a bx

y

o

Figure 3

XiXi-1

S1 S2

Sn

R2

y=f(x)

Ri

a bx

y

o

Figure 4

XiXi-1

R1 Rn

ix

1x

2x nx

approximated by


Step 3: Taking limit—Notice that the approximation appears to become better and better as the strips become thinner and thinner. So we define the area of the region as the limit value (if it exists) of the sum of areas of the approximating rectangles, that is

n

iii

PxxfA

10||||

)(lim(1)

Remark 2: In step 1, we have no need to divided the interval [a,b] into n subintervals with equal length. But for purposes of calculation, it is often convenient to take a partition that divides the interval into n subintervals with equal length.(This is called a regular partition)

Remark 1: It can be shown that if f is continuous, then the limit (1) does exist.


Solution Since y=x2+1 is continuous, the limit (1) must exist for all possible partition P of the interval [a, b] as long as ||P|| 0. To simplify things let us take a regular partition. Then the partition points are

x0=0, x1=2/n, x2=4/n, … , xi=2i/n, … , xn=2n/n=2

So the norm of P is

||P||=2/n

Let us choose the point to be the right-hand endpoint:

= xi=2i/n

By definition, the area is

ix

Example 2: Find the area under the parabola y=x2+1 from 0 to 2.

ix


Example 3 Find the area under the cosine curve from 0 to b, where .2/0 b

Solution We choose a regular partition P so that

||P||=b/n

and we choose to be the right-hand endpoint of the ith sub-interval:

=xi=ib/n

Since ||P|| 0 as n , the area under the cosine curve from 0 to b is

ix

ix

314

1

22

10||||

)(lim)(lim

n

inn

i

n

n

iii

PfxxfA

Remark 3: If is chosen to be the left-hand endpoint , one will obtain the same answer.

ix


b

xxfA

nb

nbn

b

nb

n

n

inib

nb

n

n

iii

P

sinlim

coslim)(lim

2

2)1(2

sin

cossin

110||||

/section 5.2 end


5.3 The Definite Integral In Chapters 6 and 8 we will see that limit of form

n

iii

PxxfA

10||||

)(lim

occurs in a wide variety of situations not only in mathematics but also in physics, Chemistry, Biology and Economics. So it is necessary to give this type of limit a special name and notation.

1. Definition of a Definite Integral If f is a function defined on a closed interval [a, b], let P be a partition of [a, b] with partition points x0 , x1 , x2 ,…., xn , where

a=x0 < x1<x2 <…< xn=b机动目录上页下页返回结束

Choose points in [xi-1 , xi] and let and

1 iii xxx

}max{|||| ixP

ix

. Then the definite integral of f from a to b is

n

iii

P

b

a

xxfdxxf1

0||||)(lim)(

if this limit exists. If the limit does exist, then f is called integrable on the interval [a, b].

upper limit integrand Note 1:

b

a

dxxf )(

lower limit

integral sign


Note 2: is a number; it does not depend on x. In fact, we have

b

adxxf )(

b

a

b

a

b

adrrfdttfdxxf )()()(

Note 3: The sum is usually called a Riemann sum.

n

iii xxf

1

)(

Note 4: Geometric interpretations

For the special case where f(x)>0,

= the area under the graph of f from a to b.b

a

dxxf )(

In general, a definite integral can be interpreted as a difference of areas:

21)( AAdxxfb

a

x

y

o a b+ +

-

Note 5: In the case of a>b and a=b, we extend the definition of as follows:

b

a

dxxf )(

If a>b, then a

b

b

a

dxxfdxxf )()(

If a=b, then 0)( a

a

dxxf

Example 1 Express

n

iiiii

Pxxxx

1

3

0||||]sin)[(lim

as an integral on the interval [0, ].

Example 2 Evaluate the integral by interpreting in terms of areas.

3

0

)1( dxx


Solution We compute the integral as the difference of the areas of the two triangles:

5.1)1( 21

3

0

AAdxx

-11 3o

x

y

y=x-1 A1

A2

2. Existence Theorem

Theorem If f is either continuous or monotonic on [a, b], then f is integrable on [a, b]; that is , the definite integral exists.

b

a

dxxf )(

Remark 1: If f is discontinuous at some points, then might exist or it might not exist. But if f is piecewise continuous, then f is integrable.

b

a

dxxf )(


Remark 2: It can be shown that if f is integrable on [a, b], then f must be a bounded function on [a, b].

3. Integral Formulas under Regular Partition

If f is integrable on [a, b], it is often convenient to take a regular partition. Then

nab

nxxx 1

and xiaxxaxax i ,,, 10

If we choose to be the right endpoint in each subinterval, then

ix

nab

ii iaxx Since ||P||=(b-a)/n, we have ||P|| 0 as n , so the definition gives


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

b

a

n

in

abn

ab

niafdxxf

1

)(lim)(

Example 3 Express as an integral on the interval [1, 2].

n

kknn n

11

111lim

Answer: dxx2

1

1

If the purpose is to find an approximation to an integral, it is usually better to choose to be midpoint of the subinterval, which we denote by . Any Riemann sum is an approximation to an integral, but if we use midpoints and a regular partition we get the following approximation:

ix

ix


Midpoint Rule

where

and

)]()([)()( 11

n

n

ii

b

axfxfxxxfdxxf

nabx

)( 121

iii xxx

Using the Midpoint Rule with n=5 we can get an approximation of integral (see page 277).dxx

2

1

1


4. Properties of the Integral

b

a

b

c

c

a

b

a

b

a

b

a

b

z

b

a

b

a

dxxfdxxfdxxf

dxxfcdxxcf

dxxgdxxfdxxgxf

abccdx

)()()(.4

)()(.3

)()()]()([.2

)(.1

Suppose all of the following integrals exist. Then

Example 4 Using the properties above and the results

b

a

abxdx 2

22

5

4.dxx


to evaluate

Order properties of the integral Suppose the following integrals exist and a<b.

5. If f(x)>0 and a<x<b, then

6. If f(x)>g(x) for a<x<b, then

7. If m<f(x)<M for a<x<b, then

8.

b

adxxf .0)(

b

a

b

adxxgdxxf .)()(

b

aabMdxxfabm )()()(

b

a

b

adxxfdxxf )()(

Proof of Property 7 Since m<f(x)<M , Property 6 gives

b

a

b

a

b

aMdxdxxfmdx )(

Using Property 1, we obtain

b

aabMdxxfabm )()()(


Example 5 Show that .5.714

1

2 dxx

Solution Notice that

xxx 221

for 1<x<4. xx 21

5.714

1

4

1

2 xdxdxx

Example 6 Proof that .sin 3

2/

6/6

dxx

/section 4.3 end


Since |x|=x for x>0, we have

Thus, by Property 7,

4.4 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus in this section gives the precise inverse relationship between the derivative and the integral. It enables us to compute areas and integrals very easily without having to compute them as limits of sums as we did in sections 4.2 and 4.3.

1. Fundamental Theorem

For a continuous function f on [a, b], we define a new function g by

dttfxgx

a )()(

Computing the derivative of g(x) we obtain


The Fundamental Theorem of Calculus, Part 1

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

is continuous on [a, b] and differentiate on (a, b), and

bxadttfxgx

a )()(

).()( xfxg

Proof If x and x+h are in [a, b], then

and so, for h ,

(2)

0

x

a

hx

adttfdttfxghxg )()()()(

x

a

hx

x

x

adttfdttfdttf )()()(

hx

xdttf )(

dttfhx

xhhxghxg

)(1)()(


For now let us assume that h>0. Since f is continuous on [x, x+h], the Extreme Value Theorem says that there are number u and v in [x, x+h] such that f(u)=m and f(v)=M, where m and M are the absolute minimum and maximum values of f on [x, x+h].

hvfMhdttfmhhufhx

a)()()(

)()()( 1 vfdttfufhx

ah

)()( )()( vfuf hxghxg


Inequality (3) can be proved in a similar manner for the case where h<0.

(3)

Combining it with (2) gives

Since h>0, we can divide this inequality by h:

By Property 7, we have

Since f is continuous at x, and u, v lie between x and x+h, we have

)()(lim)(lim),()(lim)(lim00

xfvfvfxfufufxvhxuh

)(lim)( )()(

0xfxg h

xghxg

h


If x=a and b, then Equation (4) can be interpreted as one- side limit.

Then Theorem 2.1.8 shows that g is continuous on [a, b].

(4)

We conclude, from (3) and the Squeeze Theorem, that

Note: This Theorem can be written in Leibniz notation as

)()( xfdttfx

adxd


Roughly speaking, equation (5) says that we first integrate f and

then differentiate the result, we get back to the original function f.

(5)

Example 1 Find the derivative of the function g(x)= .10

2 dttx

Example 2 Find .sin1

1

22

dttx

dxd

Example 3 Find .1

sin2

dtx

x tt

dxd

Example 4 Find .1lim2

2

21

0dtt

h

hh

The second part of the Fundamental Theorem of Calculus provides us with a much simpler method for the evaluation of integrals.

The Fundamental Theorem of Calculus, Part 2

If f is continuous on [a, b], then

.fF

).()()( aFbFdxxfb

a


where F is any antiderivative of f, that is

Proof We know from part 1 that g is an antidervitive of f. If F is any other antidervitive of f on [a, b], then f and g differ only by a constant:

(6) F(x)=g(x)+C

for a<x<b. But both f and g are continuous on [a, b] and so, by taking limits of Equation (6)(as x a- and x b+ ), we see that it also holds when x=a and x=b.

Noticing g(a)=0 and using Eq.(8) with x=a and x=b, we have

F(b)-F(a)=[g(b)+C]-[g(a)+C]=g(b)-g(a)

= .)( dttfb

aNotation Set F(x)| =F(b)-F(a). So we haveb

a

ba

b

axFdxxf |)()(


Example 5 Evaluate the integral .1

2

3dxx

Example 6 Find the area under the cosine curve from 0 to b, where .2/0 b

2. The Indefinite Integral

Because of the relation given by the Fundamental Theorem between antidervitive and integral, the notation is traditionally used for an antidervative of f and called an indefinite integral. Thus

dxxf )(

(7) )()(')()( xfxFmeanxFdxxf Attention A definite integral is a number, whereas an indefinite integral is a function. Then the connection between them is given by Part 2 of the Fundamental Theorem, i.e.,

(7) .|)()( ba

b

adxxfdxxf

3. Table of Indefinite Integrals

CxxdxxCxxdxx

CxxdxCxxdx

CxxdxCxxdx

nCdxx

dxxgdxxfdxxgxf

dxxfcdxxcf

nxn n

csccotcscsectansec

cotcsctansec

sincoscossin

)1(,

)()()]()([

)()(

22

1

1


Example 7 Find the general indefinite integral

dxxx )sec310( 24

Example 8 Evaluate

3

41

3

1

1

13

1

13

1 2

x

dxx

Example 9 Evaluate

21,

10,)()( 2

2

0 xx

xxxfwheredxxf

Example 10 What is wrong with the following calculation?

End /section 4.4

dxxxx

)tansec1

(


4.5 The Substitution Rule

Because of the Fundamental Theorem of Calculus , we can integrate a function if we know an antiderivative. But the anti-differentiation formulas in Section 4.4 do not suffice to evaluate integrals of more complicated functions. It is necessary for us to develop some techniques to transform a given integral into one of the forms in the table.

The substitution rule drives from the Chain Rule

)('))(('))](([ xgxgFxgFdxd

CxgFdxxgxgF ))(()('))(('


Integrating it, we get

(1)

If we make the substitution u=g(x), then Eq.(1) becomes

fF ' duuFCuFCxgFdxxgxgF )(')())(()('))(('

duufdxxgxgF )()('))(('

Thus we have proved the following rule:

The Substitution Rule for Indefinite Integral

If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

duufdxxgxgf )()('))((


or, writing , we get

Example 1 Find dxxx )2cos( 43

Example 2 Evaluate dxx 43

Example 3 Evaluate dxx

x sin1cos

dxx

x2cos

Example 4 Evaluate

Example 5 Calculate dxx

x

ee 1


The Substitution Rule for Definite Integral

If g'(x) is continuous on [a, b] and f is continuous on the range of g, then

)(

)()()('))((

bg

ag

b

aduufdxxgxgf

Example 6 Evaluate .434

0dxx

Example 7 Evaluate .2

ln1 dx

e

e xx

Example 8 Evaluate .1 14

1

12 dxxx


The next theorem uses the Substitution Rule for Definite Integral to simplify the calculation of integral of functions that possess symmetry properties.

Integrals of Symmetric Functions

Suppose f is continuous on [-a, a].

(a) If f is even [f(-x)=f(x)], then

(b) If f is odd [f(-x)=-f(x)], then

.)(2)(0

aa

adxxfdxxf

.0)( dxxfa

a

Example 9 Evaluate .2/

2/ 1sin

2

2

dxx

xx

Example 10 Show that

.)(sin)(sin020

dxxfdxxxf


END


Documents

Chapter 5 Integrals 机动 目录 上页 下页 返回 结束 5.2 Area 5.3 The Definite Integral 5.4 The Fundamental Theorem of Calculus 5.5 The Substitution Rule

Chapter 5 Integrals 机动目录上页下页返回结束 5.2 Area 5.3 The Definite Integral 5.4 The Fundamental Theorem of Calculus 5.5 The Substitution Rule