TRAPEZIUM RULE

TRAPEZIUM RULE

Prepared by Mdm Nurul Aini

APPROXIMATION FOR DEFINITE INTEGRAL
(AREA UNDER THE CURVE)

There are times when we need to evaluate

the integral ( ) exactly but cannot do


so by the integration method we have learnt. = ( )

HOW TO SOLVE THIS PROBLEM?

• Since the integral ( ) gives the area


bounded by the curve = ( ), the axis

and the ordinates x=a and x=b,

• an approximate value for the integration A=

can be found by estimating this area by

another method.

Trapezium Rule

The Trapezium Rule is a method of finding the approximate value of an integral between two limits

The area involved is divided up into a
number of parallel strips of equal width, h

Each area is considered to be a
trapezium

Suppose we wish to find the area under a
curve y = f(x) and x=a and x=b

1 1 1 1 We can divide the area into 4 trapezium
( ) ≈ 2 0 + 1 ℎ + 2 1 + 2 ℎ + 2 2 + 3 ℎ + 2 3 + 4 ℎ of equal width, h. The parallel sides of
1 −
≈ 2 ℎ[ 0 + 4 +2 1 + 2 + 3 ], where ℎ= the 4 trapezium are given by the 5
ordinates, 0, 1, 2, 3 and 4

The Trapezium Rule with n intervals is given by

For n equal-width strips/sub-intervals, n = Number of sub-intervals Using more strips of smaller width
(n+1) ordinates are used. Note that the = (Number of ordinates) - 1 involves more calculations but also

widths of all strips must be equal yields a more accurate
approximation.

If a graph concave downwards
over the whole interval from
a to b, the trapezium rule

under-estimates the actual area.

If a graph concave upwards
over the whole interval from

a to b, the trapezium rule
over-estimate the actual area.

Example :

Given n = 4 , a = 1 and b = 3 Accuracy of values in the

With 4 sub-intervals, the width of table must be 1 or 2 places
each strip better than the answer’s

− accuracy. Must be consistent.
ℎ =
1 1.5 2 2.5 3
= 3−1 ln 0 0.6368 0.8323 0.9572 1.0482

4

= 0.5

ln x ≈ 1 ℎ[ 0 + 4 + 2( 1+ 2 + 3)]
2
≈ 1 (0.5)[0 + 1.0482 + 2(0.6368+0.8323+0.9572)]

2

≈ 1.475

Checkpoint 12.4 Answer : 1.310(3 dp)
1. Answer : 0.923 (3 dp)
Answer : 1.257 (3 dp)
2.
3.

Checkpoint 12.4 Answer :
4. Point of intersection : (2,6)
Area under the curve : 5.317 (3d.p)
5. Area of the shaded region : 0.683 (3s.f)

Answer :
a) 3.536 (3 d.p)
b) 4.152 (3 d.p)

THANK YOU