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C H A P TE R 0


A P R E V IE W O F
  C ALC ULUS
C O N TE N TS


 1. T h e A r e a    P r o b le m
 2 . The Ta ng       e n t P r o b le m
 3 . V e lo c it y
 4 . T h e L im it   of a
  S e que nc e
 5 . The S u m       o f a S e r ie s
1. THE AREA PROBLEM

       The area of a polygone can be
        defined by subdividing it into
                        F
        triangles A

                                  E
               B


                   C          D
Area (ABCDEF)=Area(ABF)+Area(BCF)+Area(CDF)+Area(DEF)
1. THE AREA PROBLEM
   The area of a curved figure can be defined
    as limit of areas of incribed and
    circonscribed polygones
      A1       A2      A3            A4   A5




               Area(Circle) = lim An
                              n →∞
How to compute the area under the
parabola y = x2 ?
It is approximated by the sum of the
areas of circumscribed rectangles
the approximation can be refined by
taking narrower circumscribed
rectangles
Finally the area can be computed as the limit
of the sum of the areas of circumscribed
rectangles




  This is the idea of Integral Calculus
2 . TH E TA N G E N T
PROB LEM

                           t


                 P




 The tangent t to the curve y=f(x) at point
 P is determined by its slope m
We can approximate m by the slope mP Q of a
secant P Q , where Q is a nearby point

                              t

               P (a,f(a))    Q (x,f(x))




                      f ( x) − f (a)
              mPQ   =
                           x−a
f ( x ) − f (a )
   m = lim mPQ    = lim
        Q→P         Q→P      x−a




This is the idea of Differential Calculus
3.
 V E L O C IT Y
 The distances (in feet) traveled by a car moving
 along a straight road measured at 1-second
 intervals are given by:

  t = time elapsed    0     1    2     3     4      5
         (s)
  d = distance (ft)   0     2   10    25    43    78

The average velocity in the time interval 2 ≤ t ≤ 4
is:
                   43 − 10
                v=         = 16.5 ft / s
                    4−2
The average velocity in the time interval 2 ≤ t ≤ 3
 is:
                         25 − 10
                    va =         = 15 ft / s
                          3− 2
The distances traveled by the car measured at 0.1-
second intervals are given by:

        t     2.0    2.1     2.2     2.3       2.4   2.5
        d  10.0 11.0 12.16 13.45 14.96 16.80
            0      2
  The average velocity in the time interval 2 ≤ t ≤ 2.5
  is:
                 16.80 − 10.00
            va =               = 13.6 ft / s
                    2.5 − 2
The average velocity computed in various interval
  is:

     t     [2 3]   [2 2.5] [2 2.4] [2 2.3]   [2 2.2]   [2 2.1]
     va    15.0     13.6    12.4    11.5      10.8      10.2

The instantaneous velocity at t = 2 is the limit of the
average velocity over smaller and smaller interval [2
t]:
                            f (t ) − f (2)
                   v = lim
                       t →2      t −2
   where f(t) is the distance (in feet) traveled
   after t seconds
4 . L IM IT O F A
 S EQUENC E

The real number π is the limit of the sequence:
a1=3.1, a2=3.14, a3=3.141, a4=3.1415, a5=3.14159,
…
                 π = lim an
                     n →∞




            1
       lim = 0
       n →∞ n
5 . TH E S U M O F A
S E R IE S

The n-digit decimal representation of a rational
number 0.d1 d2 d3 … dn is the finite sum:

                      d1  d2   d3       dn
 0.d1d 2 d 3 ...d n =    + 2 + 3 +... + n
                      10 10   10       10

The infinite decimal representation of a real
number 0.d1 d2 d3 … dn … is the infinite sum:
                        d1  d2   d3
      0.d1d 2 d 3 ... =    + 2 + 3 +...
                        10 10   10
Therefore the infinite decimal representation 0.333 …
is:
        1                3     3     3
          = 0.333... =     + 2 + 3 +...
        3               10 10       10
  Similarly we can compute the infinite
  sum:
  1 1 1        1
   + + + ... + n + ... As the limit of the sequence:
  2   4   8      2
     1     1 1           1 1          1
s1 = , s2 = + ,..., s10 = + + ... +      = 0.99902344,...,
     2     2 4           2 4        1024
      1 1       1                        1 1 1           1
s16 = + + ... + 16 = 0.99998474,..., sn = + + + ... + n ,...
      2 4      10                        2 4 8          2
              1 1 1       1
               + + +... + n +... = lim sn =1
              2 4 8      2         n→∞

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Calculus 1

  • 1. C H A P TE R 0 A P R E V IE W O F C ALC ULUS
  • 2. C O N TE N TS  1. T h e A r e a P r o b le m  2 . The Ta ng e n t P r o b le m  3 . V e lo c it y  4 . T h e L im it of a S e que nc e  5 . The S u m o f a S e r ie s
  • 3. 1. THE AREA PROBLEM  The area of a polygone can be defined by subdividing it into F triangles A E B C D Area (ABCDEF)=Area(ABF)+Area(BCF)+Area(CDF)+Area(DEF)
  • 4. 1. THE AREA PROBLEM  The area of a curved figure can be defined as limit of areas of incribed and circonscribed polygones A1 A2 A3 A4 A5 Area(Circle) = lim An n →∞
  • 5. How to compute the area under the parabola y = x2 ?
  • 6. It is approximated by the sum of the areas of circumscribed rectangles
  • 7. the approximation can be refined by taking narrower circumscribed rectangles
  • 8. Finally the area can be computed as the limit of the sum of the areas of circumscribed rectangles This is the idea of Integral Calculus
  • 9. 2 . TH E TA N G E N T PROB LEM t P The tangent t to the curve y=f(x) at point P is determined by its slope m
  • 10. We can approximate m by the slope mP Q of a secant P Q , where Q is a nearby point t P (a,f(a)) Q (x,f(x)) f ( x) − f (a) mPQ = x−a
  • 11. f ( x ) − f (a ) m = lim mPQ = lim Q→P Q→P x−a This is the idea of Differential Calculus
  • 12. 3. V E L O C IT Y The distances (in feet) traveled by a car moving along a straight road measured at 1-second intervals are given by: t = time elapsed 0 1 2 3 4 5 (s) d = distance (ft) 0 2 10 25 43 78 The average velocity in the time interval 2 ≤ t ≤ 4 is: 43 − 10 v= = 16.5 ft / s 4−2
  • 13. The average velocity in the time interval 2 ≤ t ≤ 3 is: 25 − 10 va = = 15 ft / s 3− 2 The distances traveled by the car measured at 0.1- second intervals are given by: t 2.0 2.1 2.2 2.3 2.4 2.5 d 10.0 11.0 12.16 13.45 14.96 16.80 0 2 The average velocity in the time interval 2 ≤ t ≤ 2.5 is: 16.80 − 10.00 va = = 13.6 ft / s 2.5 − 2
  • 14. The average velocity computed in various interval is: t [2 3] [2 2.5] [2 2.4] [2 2.3] [2 2.2] [2 2.1] va 15.0 13.6 12.4 11.5 10.8 10.2 The instantaneous velocity at t = 2 is the limit of the average velocity over smaller and smaller interval [2 t]: f (t ) − f (2) v = lim t →2 t −2 where f(t) is the distance (in feet) traveled after t seconds
  • 15. 4 . L IM IT O F A S EQUENC E The real number π is the limit of the sequence: a1=3.1, a2=3.14, a3=3.141, a4=3.1415, a5=3.14159, … π = lim an n →∞ 1 lim = 0 n →∞ n
  • 16. 5 . TH E S U M O F A S E R IE S The n-digit decimal representation of a rational number 0.d1 d2 d3 … dn is the finite sum: d1 d2 d3 dn 0.d1d 2 d 3 ...d n = + 2 + 3 +... + n 10 10 10 10 The infinite decimal representation of a real number 0.d1 d2 d3 … dn … is the infinite sum: d1 d2 d3 0.d1d 2 d 3 ... = + 2 + 3 +... 10 10 10
  • 17. Therefore the infinite decimal representation 0.333 … is: 1 3 3 3 = 0.333... = + 2 + 3 +... 3 10 10 10 Similarly we can compute the infinite sum: 1 1 1 1 + + + ... + n + ... As the limit of the sequence: 2 4 8 2 1 1 1 1 1 1 s1 = , s2 = + ,..., s10 = + + ... + = 0.99902344,..., 2 2 4 2 4 1024 1 1 1 1 1 1 1 s16 = + + ... + 16 = 0.99998474,..., sn = + + + ... + n ,... 2 4 10 2 4 8 2 1 1 1 1 + + +... + n +... = lim sn =1 2 4 8 2 n→∞