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.
∫ 2
0
3
1 +t^2
dt (Use 8 intervals) [3.323]
2.
∫ π
2
0
1
1 +sinθ
dθ (Use 6 intervals) [0.997]
3.
∫ 3
1
lnx
x
dx (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
3
cd^3 or
1
3
d( 6 a+ 2 cd^2 ) (3)
y
y 1 y 2 y 3
y abxcx^2
d O d x
Figure 45.3
Since 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
∫b
a
ydx
≈
1
3
d(y 1 + 4 y 2 +y 3 )+
1
3
d(y 3 + 4 y 4 +y 5 )
+
1
3
d(y 2 n− 1 + 4 y 2 n+y 2 n+ 1 )