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.