Trapezium Rule
The trapezium rule is a method of finding the
Approximate integration of a function.
Example

Evaluate
3
x 4 dx
2
3
5   5 



x
3
2
4



x dx           42.2
2
 5 2  5   5 

3
5
In general, when the graph of y = f(x) is approximately

b
linear for a ≤ x ≤ b, the value of the integral
f x dx can
a
be approximated by the area of a trapezium. This can be
seen easily from the diagram below.
y  f x 
Area of trapezium
=
a
b
1
b  a  f a   f b
2
For the previous problem
1
Area of trapezium  3  2 f 3  f 2
2
1
 81  16  48.5
2
Which is close but not close enough
By increasing the number of trapeziums between a and b
we obtain a more accurate approximation. The increased
number of trapeziums gives rise to the following formula

b
1
f x dx  d  y0  yn1  2 y1  y2  y3  ...
2
a
Or in other words…

b
1
f x dx  d
2
a
(sum of end + twice sum of
the rest)
points
Example
Use the trapezium rule with 5 ordinates to find the
approximate value of

3
1
1
1 x2
dx
, giving your
answer to three decimal places.
x0 = 1
y0 = 0.707...
x1 = 1.5
y1 = 0.554...
x2 = 2
y2 = 0.447...
x3 = 2.5
y3 = 0.371...
x4 = 3
y4 = 0.316...
Difference between each x
coordinate worked out by going from
x = 1 to x = 3 in five ordinates.
Put x = 1 into

3
1
1
1 x2
dx 
1
1 x2
1
 0.50.707  0.316  20.554  0.447  0.371
2
 0.942
Example
By considering four strips of equal width and considering the
approximate area to be that of four trapeziums, estimate
the value of

1
cos x dx
, giving your answer to 4 d.p.
0
x0 = 0
For problems involving trig functions,
calculators must be in radian mode
y0 = 1
x1 = 0.25 y1 = 0.877...
x2 = 0.5
y2 = 0.760...
x3 = 0.75 y3 = 0.647...
x4 = 1
y4 = 0.540...

1
0
cos x dx 
1
 0.251  0.540  20.877  0.760  0.647 
2
 0.7640
Example
Use the trapezium rule, with five ordinates, to evaluate

0.8
x2
e dx
, giving your answer to four decimal places
0
x0 = 0
y0 = 1
x1 = 0.2
y1 = 1.0408...
x2 = 0.4
y2 = 1.1735...
x3 = 0.6
y3 = 1.4333...
x4 = 0.8
y4 = 1.8964...
Put x = 0 into

0.8
x2
0
e dx 
e
x2
1
 0.21  1.8964  21.0408  1.1735  1.433
2
 1.0192