2413 Calculus I
Chapter 5(1)
Velocity & Distance
Sigma Notation
Reimann Sums
Given the graph for velocity, find the distance traveled in 10 minutes
5
4
Velocity
(ft/min)
3
2
1
1
2
3
4
5
6
7
8
9
10
Time
First 3 minutes: D = 3(1) = 3 ft.
Second 3 minutes: D = 3(5) = 15 ft.
Last 4 minutes: D = 4(3) = 12 ft.
Total Distance: 30 ft.
Observation:
The distance is the
area under the
velocity curve
Sigma Notation
7
 3i  2
i 2
This is the sum of the
answers that you get when
each integer from 2 through 7
is put in the function
3(2)  2  3(3)  2  3(4)  2 
3(5)  2  3(6)  2  3(7)  2
 93
Formulas for Summation
n(n  1)
i

2
i 1
n
n
 c  cn
i 1
n(n  1)(2n  1)
i 

6
i 1
n
2
n (n  1)
i 

4
i 1
2
n
3
2
Example 1
i 1

2
i 1 n
n
1
2
n
Move anything out
that is not an i
n
 i 1
Use the formulas
i 1
1  n(n  1)

 1n 
2 
n  2

n3

2n
Example 2
A man gives his son and allowance of:
Jan 1: 3 cents
Jan 2: 12 cents
Jan 3: 27 cents
and continues 3(day2) for the rest of the year.
How much does he give his son by day 365?
365
 3i
i 1
2
 n(n  1)(2n  1) 
3
 and n  365
6


$162,757.15
Example 3
n
4i
Lim  2
n 
i 1 n
1
Lim 2
n  n
Move anything out
that is not an i
n
 4i
Use the formulas
i 1
1  n(n  1) 
4n 2  4n
4
Lim 2  4

Lim
 2

2
n  n
n 
2 

2
2n
Graph y = x2 + 3 on [0,2] and approximate the area
under the curve with 4 rectangles that touch the curve
on the left side.
Area  Height Width
 ( x 2  3)( 12 )
(0  3)( 12 )  3 2
( 14  3)( 12 )  13 8
(1  3)(12 )  2
1
2
1
2
1
2
1
2
1
1
2
3
2
2
( 94  3)( 12 )  218
Area  31 4
This is known as a Lower/Left Riemann Sum
Graph y = x2 + 3 on [0,2] and approximate the
area under the curve with 4 rectangles that
touch the curve on the right side.
( 14  3)( 12 )  13 8
(1  3)( 12 )  2
( 94  3)( 12 )  218
1
2
1
2
1
2
1
2
1
1
2
3
2
2
(4  3)( 12 )  7 2
Area  39 4
This is known as an Upper/Right Riemann Sum
The actual area will be near the middle of the two.
To find the actual area you need an infinite number of
rectangles.
Look for a pattern from
our rectangles
( 14  3)( 12 )  13 8
1
2
1
2
1
2
1
2
1
(1  3)( 12 )  2
1
2
3
2
2
(height )( width)
n
units
Lim  ( function)(
)
n 
# of rect.
i 1
1
2
x-values used were:
this is from:
1(
2units
4rect
),2(
3
2
2
2
, , ,
4
2
2units
4rect
2units
4 rect
),3(
),4(
2units
4 rect
)
i(2 units) 2i

n rect.
n
Translate to n rectangles:
n
Put into infinite limit:
Lim 
n 
2
n

Lim  (( 2ni )2  2)( n2 )
n 
4 n ( n 1)(2 n 1)
6
n2
i 1
 2n

26

3
Actual area
under curve