1. TOPIC : 7
NUMERICAL METHOD

2. LECTURE 1 OF 4
7.0 NUMERICAL METHOD
7.1 The Trapezium Rule

3. OBJECTIVES Derive the trapezium rule by dividing the
area represented by into n
trapezium each with width h.
b) Use the the rule to approximate

4. Numerical Method can be used to find
(a) the approximate values for definite integrals .
Example:
(b) the approximate solutions for non- linear equations.
Example:

5. THREE TYPES OF NUMERICAL METHODS
TRAPEZIUM RULE
( to find the approximate values of definite integrals )
ITERATION METHOD
NEWTON RAPHSON METHOD
( (ii) and (iii) - to find the approximate solutions for non-linear equations )

6. THE TRAPEZIUM RULE The area between y =f(x), the x-axis and
the line x=a and x=b.

7. NOTE : 5 ordinates => x0 , x1 , x2 , x3 and x4
Thus n = 4
This method divides
the area under the
curve y = f(x) into
n vertical strips.
The width of each
strip is h

8. = (sum of parallel sides x width)
Area of trapezium:
= (sum of parallel sides x width)
The total area the sum of trapezium area
under the curve b a h

9. Therefore The Trapezium Rule is :
n 7
Where h =

10. USING THE CALCULATOR Key in MODE 1 (COMP)
EX: EVALUATE
Key in
MODE 1 (COMP)
ALPHA Y = 1 ÷ ( 1+ ALPHA X X2)
CALC (X?)
1 = (0.5)
1.5 = ( )
CONTINUE THE PROCESS.

11. Example 1
Evaluate
using 6 strips by the trapezium
rule.

12. Solution The integration interval (b-a) = 4 – 1 = 3 units So h =
The values of x at which y is calculated are
1, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0
Construct a table :

13. X
Y = 1/1+x2
X0 = 1
X1 = 1.5
X2 = 2.0
X3 = 2.5
X4 = 3.0
X5 = 3.5
X6 = 4.0
Y0=0.5
Y6=
Y1=
Y2=
Y3=
Y4=
Y5=
Totals

14. thus

15. Using 6 strips by the trapezium rule.
EXAMPLE 2
Evaluate
Using 6 strips by the trapezium rule.
Solution
h =

16. X Y=
X0 = 0
X1 = 0.1
X2 = 0.2
X3 = 0.3
X4 = 0.4
X5 = 0.5
X6 = 0.6
Y0=1
Y6=1.25
Y1=
Y2=
Y3=
Y4=
Y5=
Totals

17. Therefore,

18. Example 3 Find an approximate value for
using trapezium rule with 5 ordinates,
giving your answer correct to 3 d.p.
NOTE : 5 ordinates => x0 , x1 , x2 , x3 and x4
Thus n = 4

19. Solution X Y = X0 = 0 X1 = X2 = X3 = X4 = Y0 = 0 Y4 = 0 Y1 = 0.84089
REMARKS :Use mod radian
h = = X
Y =
X0 = 0
X1 =
X2 =
X3 =
X4 =
Y0 = 0
Y4 = 0
Y1 =
Y2 = 1.000
Y3 =
Totals

20. By the Trapezium Rule :

21. Example 4 Use the trapezium rule, with 5 ordinates, to evaluate
Correct your answer to 3 decimal places.

22. Solution: n=4 So, h = 0.8/4 = 0.2 X Y = X0 = 0 X1 = 0.2 X2 = 0.4
Totals
2.8965
3.6476

23. Therefore;