Transcript Area

Area
Sigma Notation, Upper and Lower Sums
Sigma Notation
 Definition – a concise notation for sums.
 This notation is called sigma notation because it uses the
uppercase Greek letter sigma, written as ∑.
 The sum of n terms a1, a2 , a3 , . . . an is written as
n
a  a
i 1
1
2
 a3  . . . .  an
where i is the index of summation, a i is the ith term of the sum,
and the upper and lower bounds of summation are n and 1.
Examples of Sigma Notation
5
 (k  1)(k  3)
k 2
5
2
k
  2k  3
k 2
  (2) 2  2(2)  3   (3) 2  2(3)  3

  (4)  2(4)  3   5   2(5)  3
2
 3  0  5  12  14
2

Examples of Sigma Notation
5
1

j 3 j
1 1 1
  
3 4 5
20 15 12 47

 

60 60 60 60
Examples of Sigma Notation
4
2
3
(
i

1)

(
i

1)

i 1
2
3
 (1  1) 2  (1  1)3    (2  1) 2  (2  1)3    3  1   3  1 


 4  12  (4  1)3 


  0  8  1  27    4  64 9  125
 8  28  68  134
 238
Summation Formulas
Using Formulas to Evaluate a Sum
 Evaluate the following summation for n = 10, 100, 1000 and
10,000.
i 1 1
 2

2
n
i 1 n
n
n
 i 1
(the index of summation is i )
i 1
n
1  n

 2   i  1
n  i 1
i 1 
1  n(n  1)

 2
 1(n) 
n  2

1  n 2  n  2n 
 2

n 
2

1  n 2  3n  n(n  3) n  3
 2



2
n  2 
2n
2n
Using Formulas to Evaluate a Sum
 Now we have to substitute 10, 100, 1000, and 10,000 in for




n.
n = 10
n = 100
n = 1000
n = 10,000
the answer is 0.65000
the answer is 0.51500
the answer is 0.50150
the answer is 0. 50015
 What does the answer appear to approach as the n’s get
larger and larger (limit as n approaches infinity)?
Area
 Finding the area of a polygon is simple because any plane
figure with edges can be broken into rectangles and triangles.
 Finding the area of a circular object or curve is not so easy.
 In order to find the area, we break the figure into rectangles.
The more rectangles, the more accurate the area will be.
Approximating the Area of a Plane
Region
 Use five rectangles to find two approximations of the area of
the region lying between the graph of
f ( x)   x 2  5
 and the x-axis between the graph of x = 0 and x = 2.
Steps
 1. Draw the graph
 2. Find the width of each rectangle by taking the larger
number and subtracting the smaller number. Then divide by
the number of rectangles designated.
 3. Now find the height by putting the x values found in
number 2 into the equation.
 4. Multiply the length times the height (to find the area of
each rectangle).
 5. Add each of these together to find the total area.
Approximating the Area of a Plane
Region
20
To find the width
 x. We now have to know
5
the height of each rectangle. To find this, we need to
find f (xi ) where i is 1, 2, 3, 4, and 5 (the number of
2 
rectangles). Therefore, we need to find f  (1)  ,
5 
2  2  2 
2 
f  (2)  , f  (3)  , f  (4)  , and f  (5)  .
5  5  5 
5 
Approximating the Area of a Plane
Region
 2   2 4   4 6  6 8 8 
0, 5  ,  5 , 5  ,  5 , 5  ,  5 , 5  ,  5 , 2
The right endpoints are the numbers on the right. (These are not
ordered pairs)
The sum of the areas of the five rectangles is:
Approximating the Area of a Plane
Region
2

  2  5  4i 2
 2 
 2  2 
2 
f  i      i   5    
 5 

  5  i 1  25
 5   5  i 1   5 
i 1
 5 

 8  5 2 5
 8   5(5  1)(10  1) 
 
i  2  

  10  6.48
6
 125  i 1
 125  

i 1
5
5
 Now let’s find the area using the left endpoints. The five
left endpoints will involve using the i – 1 rectangle. This
answer will be too large because there is lots of area
being counted that is not included (look at the graph).
Approximating the Area of a Plane
Region
 2(i  1)  2  5 
f
    
 5  5  i 1 
5

i 1
 2i  2  2  
f
  
 5  5  
 2 
  2i  2  2
  2  5   4i 2  8i  4 
  
 5 
  5      

 5 
5 
25
i 1  
  5  i 1 

5
8 5 2 16 5
242 5

i 
i
1



125 i 1
125 i 1 125 i 1
8  5(5  1)(10  1)  16  5(5  1)  242
 5




125 
6
 125  2  125
 8.08

Approximating the Area of a Plane
Region
 The true area must be somewhere between these two
numbers.
 The area would be more accurate if we used more
rectangles.
 Let’s use the program from yesterday to find the area
using 10 rectangles, 100 rectangles, and 1000 rectangles.
 What do you think the true area is?
Upper and Lower Sums
 An inscribed rectangle lies inside the ith region
 A circumscribed rectangle lies outside the ith region
 An area found using an inscribed rectangle is smaller
than the actual area
 An area found using a circumscribed rectangle is larger
than the actual area
 The sum of the areas of the inscribed rectangles is called
a lower sum.
 The sum of the areas of the circumscribed rectangles is
called an upper sum.
Example of Finding Upper and
Lower Sums
 Find the upper an lower sums for the region bounded by
the graph of
 f ( x)  x 2  2 x  1 and the x  axis between x  0 and x  2
 Remember to first draw the graph.
 Next find the width using the formula
ba 20 2
x 


n
n
n
Example of Finding Lower and
Upper Sums
Left endpoints
Right endpoints
2
mi  0  (i  1)  
n
(0 because a  0)
2
Mi  0  i  
n
Lower Sum
 2i  2  2 
s (n)   f (mi )x   f 
 
n

 n 
i 1
n 1
2
n 
 2i  2 
 2i  2    2 
 
 2
 1  



n 
 n    n 
i=1  
n
n
Example of Finding Lower Sum
 4i 2  8i  4 4i  4   2 
 

 1  
2
n
n
i 1 
 n 
8 n 2 16 n
8 n
8 n
8
 3  i  3  i  3 1  2  i  2
n i 1
n i 1
n i 1
n i 1 n
n

n
2 n
1  1

n i 1
i 1
8  n(n  1)(2n  1)  16  n(n  1)  8
8  n(n  1)  8
2


(1)


(
n
)

( n)
 3
 3

3 
2 
2
n 
6
n  2  n
n
 n  2  n
4(2n3  3n 2  n) 8n 2  8 8 4n 2  4


 3
 8n  2
3
3
2
3n
n
n
n
Example of Finding Lower Sum
Find a common denominator and combine terms:
8n  12n  4n  24n  24  24 12n  12n 24n 6n


 3
3
3
3
3n
3n
3n
3n
14n3  12n 2  8n 14 4 8

   2
3
3n
3 n 3n
3
2
3
2
3
Finding an Upper Sum
 Using right endpoints
 2i  2 
S (n)   f ( M 1 )x   f   
 n  n 
i 1
i 1
n
n
  2i  2
 2i    2  8 n 2 8
      2    1    3  i  2

n
 n    n  n i 1
i 1   n 
n
8  2n3  3n 2  n  8  n( n  1)  2
 3
 2 
  ( n)
n 
6
 n  2  n
8n3  12n 2  4n 4n 2  4n


2
3
2
3n
n
8n3  12n 2  4n  12n3  12n 2  6n3

3n3
14n3  24n 2  4n 14 8
4




3n3
3 n 3n 2
n
2 n
i  1

n i 1
i 1
Limit of the Lower and Upper Sums
 Let f be continuous and nonnegative on the interval
 [a, b]. The limits as n —›∞ of both upper and lower
sums exist and are equal to each other. That is,
lim s (n)  S (n)
n 
Definition of the Area of a Region in
the Plane
 Let f be continuous and nonnegative on the interval
 [a, b]. The area of the region bounded by the graph of f,
the x-axis, and the vertical lines x = a and x = b is
n
Area  lim  f  ci  x
n 
i 1
b  a

where x 
n
xi 1  ci  xi