Computational Physics - Department of Physics

5.1 Newton-Cotes Quadrature 111

PN(x) =

N

∑

i= 0

∏

k 6 =i

x−xk

xi−xk

yi,

we could attempt to approximate the functionf(x)with a first-order polynomial inxin the
two sub-intervalsx∈[x 0 −h,x 0 ]andx∈[x 0 ,x 0 +h]. A first order polynomial means simply that
we have for say the intervalx∈[x 0 ,x 0 +h]

f(x)≈P 1 (x) = x−x^0

(x 0 +h)−x 0

f(x 0 +h)+x−(x^0 +h)

x 0 −(x 0 +h)

f(x 0 ),

and for the intervalx∈[x 0 −h,x 0 ]

f(x)≈P 1 (x) =

x−(x 0 −h)

x 0 −(x 0 −h)

f(x 0 )+

x−x 0

(x 0 −h)−x 0

f(x 0 −h).

Having performed this subdivision and polynomial approximation, one fromx 0 −htox 0 and
the other fromx 0 tox 0 +h,
∫a+ 2 h
a

f(x)dx=

∫x 0

x 0 −h

f(x)dx+

∫x 0 +h

x 0

f(x)dx,

we can easily calculate for example the second integral as
∫x 0 +h
x 0
f(x)dx≈

∫x 0 +h

x 0

(

x−x 0

(x 0 +h)−x 0 f(x^0 +h)+

x−(x 0 +h)

x 0 −(x 0 +h)f(x^0 )

)

dx,

which can be simplified to
∫x 0 +h
x 0

f(x)dx≈

∫x 0 +h

x 0

(

x−x 0

h

f(x 0 +h)−

x−(x 0 +h)

h

f(x 0 )

)

dx,

resulting in ∫
x 0 +h
x 0

f(x)dx=

h

2

(f(x 0 +h)+f(x 0 ))+O(h^3 ).

Here we added the error made in approximating our integral with a polynomial of degree 1.
The other integral gives
∫x 0
x 0 −h

f(x)dx=

h

2

(f(x 0 )+f(x 0 −h))+O(h^3 ),

and adding up we obtain
∫x 0 +h
x 0 −h

f(x)dx=

h

2

(f(x 0 +h)+ 2 f(x 0 )+f(x 0 −h))+O(h^3 ), (5.3)

which is the well-known trapezoidal rule. Concerning the error in the approximation made,
O(h^3 ) =O((b−a)^3 /N^3 ), you should note the following.This is the local error!Since we are
splitting the integral fromatobinNpieces, we will have to perform approximatelyNsuch
operations. This means that theglobal errorgoes like≈O(h^2 ). To see that, we use the trape-
zoidal rule to compute the integral of Eq. (5.1),

I=

∫b

a

f(x)dx=h(f(a)/ 2 +f(a+h)+f(a+ 2 h)+···+f(b−h)+fb/ 2 ), (5.4)

with a global error which goes likeO(h^2 ).