Trapezoidal Rule
In the last section we looked at the 1 st Fundamental Theorem of
Calculus.
Recall: MRAM was generally more efficient in approximating
integrals rather than LRAM and RRAM
Rather than using rectangles, we can get a better approximation
using trapezoids.
b a
if we partition 
a, binto n subintervals of equal length h 
.
n
Then the region between the curve and the x-axis is approximated
by adding the trapezoids.
y0 y1
y yn
y y 2
h 1
... h n 1
2
2
2
x dx h 
f 
b
a
h
 y0 2 y1 2 y2 ... yn 
2
Remember h is just the width of each trapezoid.
Note: the trapezoid rule is equivalent to
LRAM RRAM
2
Example:
Using the trapezoid rule with n=5, determine
1
x dx.
2
0
Solution:
b a 1 0 1
h


n
5
5
h
T 
y0 2 y1 2 y2 2 y3 2 y4 y5 
2
2
2
2
2
1 2
1  2  3  4  2 

 0 2  2  2  2  1 

10 
5  5  5  5 


1
2
8 18 32
 
0     1

10  25 25 25 25 

0.34
Note: the exact value is
1
3
Note: when the curve is concave up, the trapezoid rule overestimates the area, and when concave down, the trapezoid rule
under-estimates the area.
The following example, gives an application of the trapezoid rule.
Example:
An observer measures the velocity of a submarine travelling under
the North Pole. Use the trapezoid rule to estimate the distance
traveled by the submarine during the 10 hour period.
t (h)
0
1
v
13 14
(km/h)
2
3
4
5
6
7
8
9
10
17
21
23
21
15
11
11
14
17
Solution:
10 0
h
1
10
10
1
s v 
t dt  
13 2 
14 2 
17 ... 2 
14 17 
0
2
162
The submarine traveled about 162 km.
Example
The temperature on Dec 12th in Waterloo was given by the following
h
equation: T 
h A B cos  , where T is the temperature in
12 
degrees Fahrenheit and h is the number of hours from midnight

0 h 24 .
a.
The initial temperature at midnight was 20
F , and at noon the
temperature was 5
F . Determine the values of A and B.
b.
Determine the average temperature for the first 8 hours.
c.
Use the Trapezoid Rule with 4 equal subdivisions to estimate
T H dH .
6
4
d.
Find an expression for the rate that the temperature is changing
with respect to H.
Solution:
a)
We know the following T 
0 20 and T 
125

0 
That is, T 
0 A B cos  
12 
20 A B cos 
0
20 A B

12 
And, T 
12 A B cos 

 12 
5 A B cos 

5 A B
Now solve these simultaneous equations.
20 A B
5 A B
Therefore A= 7.5and B= 12.5.
b) The average temperature is given by:
b
1
T 
h dh .
a
b a 
1 8
h

Tav 
7.5 12.5cos 
dh




8 0 0 
12 
Which via fnInt on your calculator is 12.6687F
6 4 1
c) Since h 
 , Then
4
2
h
T 
y 2 y1 2 y2 2 y3 2 y4 y5 
2 0

4 
4.5
1
7.5 12.5cos 12 2 7.5 12.5cos 12 







2



2 

5 
5.5
6
2 7.5 12.5cos 12 2 7.5 12.5cos 12 

7.5 12.5cos 
  


12 



1
 
85.5507 2 1.388
4
The average temperature is about 21.388
F.
d) We need only take the derivative of T with respect to h
d 

h  

h
7.5 12.5cos 
12.5sin 
  
 

dh 
12  12
12 
Example
A pond in your backyard has an average depth of 2m. using a scaled
map, you measure the distances across the pond at 10m intervals
12m
34m
55m
77m
45m
10m
36m
8m
Using the Trapezoid Rule, estimate the volume of the pond.
Solution:
First remember we are finding the volume (length x width x
depth).
Now the average depth is 2m (which is the h value), the width’s
area is 10m, and the lengths are given on the diagram.
The Trapezoid Rule will give the pond’s surface area approximation,
we need only multiply this by the average depth to obtain the
volume.
V Trapezoid Area depth
10

 
1 
0 2
12 2 
34 2 
55 ... 2 
8 1
0
2


2

2670 
2
5340
The volume is approximation 5340 m3.