ABSTRACT
Let the range of integration be divided into n equal intervals each of width d, such that nd = b − a, i.e. d = b − a
n The ordinates are labelled y1, y2, y3, . . . , yn+1 as
shown. An approximation to the area under the curve
may be determined by joining the tops of the ordinates by straight lines. Each interval is thus a trapezium, and since the area of a trapezium is given by:
area = 1 2
(sum of parallel sides) (perpendicular distance between them) then
y dx ≈ 1 2
(y1 + y2)d + 12(y2 + y3)d
+ 1 2
(y3 + y4)d + · · · 12(yn + yn+1)d
≈ d [
1 2
y1 + y2 + y3 + y4 + · · · + yn
+ 1 2
yn+1 ]
i.e. the trapezoidal rule states:
a y dx ≈
( width of interval
){ 1 2
( first + last ordinate
)
+ (
sum of remaining ordinates
)} (1)
Problem 1. (a) Use integration to evaluate, correct to 3 decimal places,
2√ x
dx (b) Use the trapezoidal rule with 4 intervals to evaluate the integral in part (a), correct to 3 decimal places.
