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Module 2

Lesson 8

Numerical Integration

Trapezoidal Rule

Edgar M. Adina

Instructor

CE50P- 2

Numerical Solutions to Engineering Problems

Lesson Objectives:

At the end of this lesson, students should be able to:

1.Explain the need for numerical techniques of integration,

2.derive the multiple-segment trapezoidal rule of

integration, and

3.use the multiple-segment trapezoidal rule of integration to

solve related applications of definite integrals.

Trapezoidal Rule

One way to approximate a definite integral is to

use n trapezoids.

In the development of this method,

assume that f is continuous and

positive on the interval [ a , b ].

So, the definite integral

represents the area of the region

bounded by the graph of f and the

x- axis, from x = a to x = b.

Trapezoidal Rule: n = 4

A 1 A 2 A 3 A 4

f ( x 0 )

f ( x 1 ) f ( x 2 )

f ( x 3 )

f ( x 4 )

a b

𝒘𝒊𝒅𝒕𝒉 =

𝒃 − 𝒂

𝒏

= 𝚫𝒙

𝚫𝒙

+

_________________________

𝑨𝟏 =

𝒃 − 𝒂

𝒏

𝒇 𝒙𝟎 + 𝒇 𝒙𝟏

𝟐

𝑨𝟐 =

𝒃 − 𝒂

𝒏

𝒇 𝒙𝟏 + 𝒇 𝒙𝟐

𝟐

𝑨𝟑 =

𝒃 − 𝒂

𝒏

𝒇 𝒙𝟐 + 𝒇 𝒙𝟑

𝟐

𝑨𝟒 =

𝒃 − 𝒂

𝒏

𝒇 𝒙𝟑 + 𝒇 𝒙𝟒

𝟐

𝑨𝑻 =

𝒃 − 𝒂

𝟐𝒏

𝒇 𝒙𝟎 + 𝟐𝒇 𝒙𝟏 + 𝟐𝒇 𝒙𝟐 + 𝟐𝒇 𝒙𝟑 + 𝒇 𝒙𝟒

• Example

Example 2

The vertical distance covered by a rocket from t=8 to t=30 seconds is given by:

  

 



 −  



 



−

=

30

8

9 8

140000 2100

140000 2000_. t dt_

t

x ln

a) Use single segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).

c) Find the absolute relative true error, for part (a).

Et

 a

Solution

b) The exact value of the above integral is

  

 



 −  



 



−

=

30

8

9 8

140000 2100

140000 2000_. t dt_

t

x ln =^11061 m

Et = TrueValue − Approximate Value = 11061 − 11868 = − 807 m

c) The absolute relative true error,  t , would be

100 11061

11061 11868 

−  t = = 7_._ 2959 %

Example 3

Answer Example 2 with n=2.

Solution

 



 















− = (^) 

−

=

f(a) f(a ih) f(b ) n

b a I

n

i

1

1

2 2

n = 2 a = 8 b = 30

2

30 − 8

n

b a h

− = = 11

 



 















− = (^) 

−

=

f( ) f(a ih) f( ) ( )

I i

8 2 30 2 2

30 8 2 1

1

 (^) f ( 8 ) 2 f( 19 ) f( 30 )  4

22 = + +

 177 27 2 48475 90167  4

22 =. + (. ) +. = 11266 m

Solution

Table 1 gives the values

obtained using multiple

segment Trapezoidal rule for

n Value Et

1 11868 - 807 7.296 ---

2 11266 - 205 1.853 5.

3 11153 - 91.4 0.8265 1.

4 11113 - 51.5 0.4655 0.

5 11094 - 33.0 0.2981 0.

6 11084 - 22.9 0.2070 0.

7 11078 - 16.8 0.1521 0.

8 11074 - 12.9 0.1165 0.

  

  

 −  

  



−

=

30

8

98 140000 2100

140000 2000_. t dt t_

x ln

Table 1: Trapezoidal Rule Values for n=1,2,3,…,

 t %  a %

Exact Value=11061 m