Numerical integration 439
Now try the following exercise
Exercise 174 Further problemson the
mid-ordinate rule
In Problems 1 to 4, evaluate the definite integrals
using themid-ordinate rule, giving the answers
correct to 3 decimal places.1.∫ 203
1 +t^2dt (Use 8 intervals) [3.323]2.∫ π
2
01
1 +sinθdθ (Use 6 intervals) [0.997]3.∫ 31lnx
xdx (Use 10 intervals) [0.605]4.∫ π
3
0√
(cos^3 x)dx (Use 6 intervals) [0.799]45.4 Simpson’srule
The approximation made with the trapezoidal rule is to
join the top of two successive ordinates by a straight
line, i.e. by using a linear approximation of the form
a+bx. With Simpson’s rule, the approximation made
is to join the tops of three successive ordinates by a
parabola, i.e. by using a quadratic approximation of the
forma+bx+cx^2.
Figure 45.3 shows a parabolay=a+bx+cx^2 with
ordinates y 1 ,y 2 and y 3 at x=−d,x=0andx=d
respectively.
Thus the width of each of the two intervals isd.The
area enclosed by the parabola, thex-axis and ordinates
x=−dandx=dis given by:
∫d
−d(a+bx+cx^2 )dx=[
ax+bx^2
2+cx^3
3]d−d=(
ad+bd^2
2+cd^3
3)−(
−ad+bd^2
2−cd^3
3)= 2 ad+2
3cd^3 or1
3d( 6 a+ 2 cd^2 ) (3)yy 1 y 2 y 3y abxcx^2d O d xFigure 45.3Since y=a+bx+cx^2 ,
at x=−d,y 1 =a−bd+cd^2
at x= 0 ,y 2 =a
and at x=d,y 3 =a+bd+cd^2
Hence y 1 +y 3 = 2 a+ 2 cd^2
And y 1 + 4 y 2 +y 3 = 6 a+ 2 cd^2 (4)Thus the area under the parabola between x=−d
and x=d in Fig. 45.3 may be expressed as
1
3 d(y^1 +^4 y^2 +y^3 ), from equations (3) and (4), and the
result is seen to be independent of the position of the
origin.
Let a definite integral be denoted by∫b
aydx and
represented by the area under the graph ofy=f(x)
betweenthelimitsx=aandx=b,asshowninFig.45.4.
The range of integration,b−a, is divided into aneven
number of intervals, say 2n, each of widthd.
Since an even number of intervals is specified, an odd
number of ordinates, 2n+1, exists. Let an approxima-
tiontothe curve over the first twointervalsbe a parabola
of the formy=a+bx+cx^2 which passes through the
tops of the three ordinatesy 1 ,y 2 andy 3. Similarly, let
an approximation to the curve over the next two inter-
vals be the parabola which passes through the tops of
the ordinatesy 3 ,y 4 andy 5 , and so on.Then∫baydx≈1
3d(y 1 + 4 y 2 +y 3 )+1
3d(y 3 + 4 y 4 +y 5 )+1
3d(y 2 n− 1 + 4 y 2 n+y 2 n+ 1 )