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By (1.11), using c
D
.1
C
1:5/=2
D
1:25,wehave
Z
1:5
1
4
1
sin.x/ dx
2
Œsin.1/
C
2 sin.1:25/
C
sin.1:5/
0:4671:
1
The relative error of this approximation is
0:4696
0:4671
0:4696
100%
0:53%;
which, as we expected, is significantly smaller than the error obtained by using only
one trapezoid.
n
1.3.3
Approximating the Integral Using
Trapezoids
It is not hard to realize that we can proceed further and split the interval into n parts,
where n
1 is an integer. Let
b
a
n
h
D
and define
x
i
D
a
C
ih
for i
D
0;1;:::n. These points
a
D
x
0
<x
1
<
<x
n1
<x
n
D
b
divide the interval from a to b into n subintervals, see Fig.
1.4
.
Due to the additive property of the integral, we have
Z
b
f.x/dx
D
Z
x
1
x
0
f.x/dx
C
Z
x
2
x
1
f.x/dx
CC
Z
x
n
x
n1
f.x/dx
a
Z
x
i C1
D
n
X
i D0
f.x/dx:
(1.13)
x
i
Let us consider one of these intervals and use the approximation derived above. If
we use x
i
and x
i C1
as the two end-points in (1.10), we get
h
x
D
D
a
x
0
x
1
x
2
x
3
x
n
b
Fig. 1.4
The interval from x
D
a to x
D
b is divided into sub-intervals of length x
i
C
1
x
i
D
h
