Finding the area
of curved
irregular shapes
LEAVING CERTIFCATE ORDINARY & HIGHER LEVELS
Find the area of your shape
Work as
accurately as
you can in
centimetres!
1A
You found the area of an A4 page!
1B
1A
Trapezoidal Rule
(a)
(b)
(c)
(d)
2.4 m
2m
Find the area of the stained glass
window.
Draw a scaled diagram of the window
using 1 m = 5 cm on square paper.
Divide the shape up into:
(i) 4 trapezium's
(ii) 5 trapezium’s
(ii) 10 trapeziums
and calculate the area of the window
each time, (using the trapezium’s).
Which part produced the smallest
error and why?
𝜋 1
𝐴 = 2 × 2.4 +
2
2
= 6.37 𝑚2
𝐴 ≈ 6.32
6.17 𝑚2
6.22
6.37
6.37−−6.32
6.17
6.22
%
%Error
Error==
××100
100==0.79%
3.14%
1.5%
6.37
6.37
Formula and Tables Page 12
What do all the terms in the formula mean?
Derivation of the formula
𝐴 = 𝑇1 + 𝑇2 + 𝑇3 + 𝑇4 + ⋯ + 𝑇𝑛
1
1
1
1
1
𝐴 = 𝑦1 + 𝑦2 ℎ + 𝑦2 + 𝑦3 ℎ + 𝑦3 +𝑦4 ℎ + 𝑦4 +𝑦5 ℎ + ⋯ + 𝑦𝑛−1 +𝑦𝑛 ℎ
2
2
2
2
2
𝐴=
ℎ
𝑦 +𝑦2 +𝑦2 +𝑦3 +𝑦3 +𝑦4 +𝑦4 +𝑦5 + ⋯ + 𝑦𝑛−1 +𝑦𝑛
2 1
ℎ
𝐴 = 𝑦1 +𝑦𝑛 + 2(𝑦2 +𝑦3 +𝑦4 +𝑦5 + ⋯ + 𝑦𝑛−1 )
2
𝒂
𝒉
𝒃
𝐴=
1
𝑎+𝑏 ℎ
2
Trapezium Rule
Why Trapezoidal Rule rather than Simpson’s Rule?
1.
Easier to derive and understand what is going on.
2.
Area of a trapezium is on Strand 3 for Ordinary Level students
from 2015.
(N.B. It is on for Foundation Level in 2014)
3.
For many practical cases it (the Trapezoidal rule) is accurate
enough.
2002
2013 HL
Using 5 intervals approximate the following integral
𝐴≈
A
ℎ
𝑦 + 𝑦𝑛 + 2 𝑦2 + 𝑦3 + 𝑦4 + ⋯ + 𝑦𝑛−1
2 1
h
y 1  y6  2( y2  y3  y4  y5 )

2
y1  1  (0)2  1
y2  1  (0.2)2  0.9797
y3  1  (0.4)2  0.9165
y4  1  (0.6)2  0.8
yn1  y5  1  (0.8)2  0.6
yn  y6  1  (1)2  0

1

1
0
0
1  x 2 dx 
0.2
1  0  2  0.9797  0.9165  0.8  0.6  
2 
1  x 2 dx  0.75924
1
0
1 − 𝑥 2 𝑑𝑥.
Other Methods
1
1 − 𝑥 2 𝑑𝑥
0
𝐴𝑟𝑒𝑎 =
1
0
𝑦 𝑑𝑥
𝑦 = 1 − 𝑥2
𝑦2 = 1 − 𝑥2
𝑥2 + 𝑦2 = 1
𝜋
𝐴𝑟𝑒𝑎 = = 0.7853
4
1 − 𝑥 2 𝑑𝑥
𝑥 = sin 𝜃
𝑑𝑥
= cos 𝜃
𝑑𝜃
1
(1 + cos 2𝜃)𝑑𝜃
2