TOPIC : 7 NUMERICAL METHOD

LECTURE 1 OF 4 7.0 NUMERICAL METHOD 7.1 The Trapezium Rule

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

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

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 )

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

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

= (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

Therefore The Trapezium Rule is : n 7 Where h =

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 = (0.30769) CONTINUE THE PROCESS.

Example 1 Evaluate using 6 strips by the trapezium rule.

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 :

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=0.05882 Y1=0.30769 Y2=0.20000 Y3=0.13793 Y4=0.10000 Y5=0.07547 Totals 0.55882 0.82109

thus

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

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=1.00504 Y2=1.02062 Y3=1.04828 Y4=1.09109 Y5=1.15470 Totals 2.25000 5.31973

Therefore,

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

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 = 0.84089 Y2 = 1.000 Y3 = 0.84089 Totals 2.68179

By the Trapezium Rule :

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

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

Therefore;