1. Finding the area under a curve using the trapezium rule
Core 2 Module
Finding the area under a curve using the trapezium rule

2. Divide area under curve into n strips (intervals) each of width h
Join tops of lines to make a series of trapeziums
Area under curve can be approximated by adding together areas of all trapeziums
y0, y1, y2 etc. are the values of f(x) and the lengths of the parallel sides of the trapeziums yn y0 y1 y2
yn-1 a b x0
x0, x1, x2 etc. are the values of x at either side of the trapeziums (ordinates)
Area of trapezium = ½(sum of parallel sides) x distance between
Area of 1st trapezium = ½(y0 + y1) x h
Total area = ½ h(y0 + y1) + ½ h(y1 + y2) + ….. + ½ h(yn-1 + yn)
= ½ h(y0 + 2y1 + 2y2 + … + 2yn-1 + yn)

3. This is called the trapezium rule
∫ f(x) dx ≈ ½ h [y0 + 2(y1 + y2 + …. + yn-1) + yn]
Where h = b – a and yr = f(xr) n b a

4. Use the trapezium rule with 4 strips to find an approximate value for the area between the curve y = xsinx, the x axis and the vertical lines x = 1 and x = 3. Give answers to 3 d.p.
4 strips = 5 ordinates
h = 3 – 1 = ½ 4
y0 = 1 x sin(1) =
y1 = 1.5 x sin(1.5) =
y2 =
y3 =
y4 = 1
1.5 2
2.5 3 3
∫ xsinx dx ≈ ½ ( ½ ) [ ( ) ]
≈ = (3 d.p.) 1

5. Group task
At a certain point in a rivers course it is 10m wide. Its depth, d metres is measured at intervals of 1m at this point.
Use the trapezium rule to find an aproximate value for the cross sectional area of the river at this point.
Draw a graph to show this information. x 1 2 3 4 5 6 7 8 9 10 d
0.93
1.26
2.81
3.64
3.87
4.09
3.92
2.36
1.68

6. Task
Trapezium rule worksheet