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					            Chapter 5               Integral

  In chapter 3 we used the tangent and velocity problems to introduce
the derivative, which is the basic idea in differential calculus.


  In this chapter, we begin with the area and the distance problems,
because they contains a basic idea in integral calculus.




                                                                  1
              Section 1 Areas and Distances
1. The Area Problem
      Find the area of the region of S that lies under the curve y=f(x)
   from a to b.
                                    y  f x
                         xa
                                        S              xb
                               a                   b




   A natural idea is that we first approximate the reign S by so many
small rectangles and then we take the Limit of the sum of these rectangle
areas as we increase the number of rectangles.                      2
Example 1    Use rectangle to estimate the area under the parabola
             y  x 2 from 0 to 1.

    Approximate each strip by a rectangle whose base is the same as the
 strip, whose height is the same as the right edge of the strip.


        n            Ln               Rn
        10       0.2850000        0.3850000
        20       0.3087500        0.3587500         o           1
                                                        R4=0.46875
        30       0.3168519        0.3501852
        40       0.3234000        0.3434000
       100       0.3283500        0.3383500
      1000       0.3328335        0.3338335
                                 1
               lim Ln  lim Rn                          L4=0.2187
               n      n      3
                                                                     3
Example 2 For the reigon S in Example 1 ,show that the sum of the
          areas of the uper approximating rectangles appracches 1/3.




                                                                 4
Apply the idea to the more general region S

1. Divide the interval [a,b] into n subintervals, and each subintervals
   has same length.  x0 , x1 ,  x1 , x2  xn1 , xn       ba
                                                           x 
                                                                 n
2. Approximate the strip Si by small rectangles
                          Si  f  xi  x

3. Make a sum       Rn  f  x1 x  f  x2 x    f  xn x

            Ln  f  x0 x  f  x1 x  f  x2 x    f  xn1 x

4.Find the limit lim Rn         lim Ln
                  n           n 


  If the two limits exist and they are the same, then we have

                         S  lim Ln  lim Rn
                             n         n 
                                                                      5
       It can be proved that, if f is continuous on [a, b], then
lim Rn and lim Ln exist,          lim Ln  lim Rn
n         n                   n     n 

   In fact, instead of using the left endpoint and right endpoint, we could
                                                                          
take the height of the ith rectangle to be the value of f at any number x i
 in the subinterval [xi-1, xi].
                  
    x1 , x 2 , , x n are called the sample points.


               n 
                     
                             
                                             
          A  lim f x1 x  f x 2 x    f x n x   
                                  
                           n
              lim  f x i x
                   n 
                          i 1




                                                                     6
Example 3 Let A be the area of the reigon that lies under the graph
          of f(x)=cosx between x=0 and x=b, where 0  b   / 2
 (a) Using right endpoints, find an expression for A as a limit. Do
     not evaluate the limit.
 (b) Estimate the area for the case b   / 2 by taking the sample
     points to be midpoints and using four subintervals.
                      M 4  1.006




                                                                      7
1. The Distance Problem
                    distance  velocity  time
  Example 4 Suppose the odometer on our car is broken and we want
            to estimate the distance driven over a 30-second time
             interval. We take speedometer readings every five second
             and record them in the following table:
        Time(s)       0      5     10     15     20     25     30
    Velocity(km/h) 27        34    38     46     51     50     45

                            1km/h=1000/3600m/s

        Time(s)       0      5     10     15     20     25     30
    Velocity(m/s)     7.5    9.4   10.6   12.8   14.2   13.9   12.5

                                                                    8
         Time(s)         0        5        10      15       20       25      30
    Velocity(km/h) 7.5            9.4      10.6    12.8     14.2     13.9    12.5

   During the first seconds the velocity doesn’t change very much, so
   we can estimate the distance traveled by assuming that the velocity
   is constant.          7.5m/s 5s  37.5m
Similarly, during the second time interval the velocity is approximately
constant and we take it to be the velocity when t=5s.
                                9.4m/s 5s  47m
  If we add similar estimates for the other time intervals, we obtain an
  estimate for the total distance traveled:
 7.5  5  9.4  5  10 .6  5  12 .8  5  14 .2  5  13 .9  5  342 m
 9.4  5  10.6  5  12.8 5  14.2  5  13.9  5  12.5 5  342m
If we had wanted a more accurate estimate, we should partition the time
Interval more brief.                                                9
                                              v
      y
                                              15
             yx   2

                                              10


                                               5
      o                    x
                       1

                                                   10           20       30   t
      L4=0.2187
                                                    L6  342m
  In general, suppose an object moves with velocity v=f(t), where
                                                          ba
1. Divide   t0 , t1 , t1, t2  tn1, tn         t 
                                                           n

2. Approximate             di  f ti  t            di  f ti 1  t
                                   n                                 n

3. Make a sum              Rn   f  ti  t         Ln   f  ti 1  t
                                  i 1                            i 1
                                         n                  n
4.Find the limit           d  lim  f  ti  t  lim  f  ti 1  t
                               n 
                                       i 1
                                                    n 
                                                           i 1
                                                                                  10
              Section 2 The Definite Integral

                     lim  f  x1  x  f  x2  x            f  xn  x 
          n
  lim              
                f x x n 
                    i
                                                                          
                                                                                  
  n 
         i 1

Except that the area and distance problems can be expressed in the form
of this type of limit of a sum, many practical problem can be considered
as this type of limit of a sum, even when f is not necessarily a positive
 function.
 Definition of A Definite Integral                     If f is a function defined for
 a  x  b , we divided the interval [a, b] into n subintervals of equal
 width x  b  a  / n . We let x0  a , x1 , x2 , , xn  b  be the endpoints of
                                                      
 these subintervals and we let        x1 , x 2 , , x n be any sample points in
 these subintervals, so x i lies in the ith subinterval x i 1 , x i  . Then the
                             

 definite integral of f from a to b is
                                                              
                                                       n
                                      f  x dx  lim  f x i x
                                  b
                              a                n
                                                      i 1
                                                                                      11
If this type of limit exits, we say that f is integrable on [a, b].
Note 1:       Symbol
                                 The function is the integrand
 Upper limit of integration
                                                                x is the Variable of Integration



                                 b
 Integral sign
                                a
                                         f (x)dx
                              Integral f from a to b

Lower limit of integration                                 When you find the value of integral,
                                                           You have evaluated the integral.
                    f  x dx   f t dt         f u du
                    b                b            b
 Note 2 :          a              a           a




                    f  x dx
                    b
  Note 3 :                        is an algebra sum of areas.                             12
                   a
Not all functions are integrable. The following theorem shows that the
most commonly occurring functions are in fact integrable.

 Theorem If f is continuous on [a, b], or if f has only a finite number
         of jump discontinuities, then f is integrable on [a, b];
         that is, the definite integral a f  x dx exists.
                                          b




 According to this theorem, if f is integrable, that means the following
type of limit exists and give the same value no matter how we choose
the sample points.
                            f  x dx  lim  f x x
                          b            n
                                            
                                                    i
                           a           n
                                             i 1


  To simplify the calculation of integral, we often take the sample points
to be right endpoints, left endpoints and even midpoints.
            f  x dx
            b

           a
                   n                   n                   n
                                                                  x  xi   
            lim  f  xi 1 x  lim  f  xi x  lim  f  i 1         x
                                                                                   13
             n 
                   i 1
                                   n 
                                         i 1
                                                      n 
                                                            i 1    2      
14
(1)

(2)

(3)



(4)

(5)

(6)
      15
(7)
                                                                                   4  3 x dx.
                                                                                   1
                                                                                             2
Example 6 Use the properties of integrals to evaluate                              0


                                                f  x dx  17 and      f  x dx  12 .
                                            10                           8
Example 7 If it is known that           0                                0

                          f  x dx.
                      10
           find   8



                                                                
                                                                    4
Example 8 Use the last property to estimate                              x dx .
                                                                 1


Example 9 Evaluate the area of the reigon bounded by
            y  x 3 and             y  x.




                                                                                             16
      Section 3 The Fundamental Theorem
                  of Calculus(TFTC)

            Differential                          Integral
             Calculus                             Calculus
                              Newton, Leibniz
          (Tangent problem)                     (Area problem)

 They discovered the closed relation between the two seeming unrelated
problems.
  TFTC gives the precise inverse relationship between the derivative
and the integral.



                                                                   17
               
 x f t dt   f  x 
 a
                           Vary upper limit function
            
             
                           ——变上限函数




                                            18
Find the derivative of the following function




                                                19
              Section 4 Indefinite Integral
  We saw that the second part of TFTC provides a very powerful
method for evaluating the denifite integral of a function.
   That is we just need to find the antiderivative of the function
then calculate the difference between the upper limit and lower
limit of integration.
Now we try to explore the methods to find antiderivatives of functions.

    In mathematic, we called all the antiderivatives of a function
indefinite integral. Denote

         f  x dx  F  x   C ,   where [F  x   C ]  f  x 

Note:                                                                 20
the difference between the definite integral and the indefinite integral.
    Function         Particular        Function       Particular
                                                     Antiderivative
                   Antiderivative
                                         cos x           sin x
  cf  x               CF  x          sin x          cos x
                                        sec2 x           tan x
f  x   g x     F  x   G x 
                                        csc2 x           cot x

  x n  1
    n                   x n 1         sec x tan x      sec x

                        n 1           csc xcot x       csc x

    The process of finding the antiderivatives is opposite to the
process of finding the derivatives.
    We can obtain the formulas for indefinite integral by inversing
those formulas.                                                  21
Table of Indefinite Integrals

 Cf  x dx  C  f  x dx    [ f  x   g x ]dx   f  x dx   g x dx
                                          x n 1
 kdx  kx  C                  x n dx 
                                          n 1
                                                 C

 sin xdx  cos x  C            cos xdx   sin x  C
  sec2 xdx  tan x  C
                                 csc2 xdx   cot x  C


   sec x tan xdx  sec x  C  csc x cot xdx   csc x  C
   Any formula can be verified by differentiating the function on the
right side and obtain the integrand.                              22
                     cos
Example 3 Evaluate  sin2  d


                       1  tan  d
                                 2
Example 4 Evaluate


                       x  2 x dx
                       2
Example 5 Evaluate
                       1




                                         23
Section 5 The Substitution Rule




                                  24
25
                             x         
                             2
Example 10 Find the value
                                   6
                                        1 dx
                            2


                            1    tan x
Example 11 Find the value                dx
                           1 1  x  x
                                   2    4




                                                26

				
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