Homework # 6
Due Monday, 02/25
1. In order to derive Simpson’s rule, we proved that, if
p
(
x
) is the unique polynomial of degree
≤
2 which
interpolates the function
f
(
x
) at
x
0
,
x
1
and
x
2
, then
x
2
x
0
p
(
x
)
dx
=
h
3
(
f
(
x
0
) + 4
f
(
x
1
) +
f
(
x
2
))
In this problem, we will do the same for the trapezoidal rule: let
p
(
x
) be the unique polynomial of
degree
≤
1 which interpolate the function
f
(
x
) at
x
0
and
x
1
. Use Lagrange’s formula to prove that:
x
1
x
0
p
(
x
)
dx
=
h
2
(
f
(
x
0
) +
f
(
x
1
))
Remark:
In class we derived this result by drawing a picture and calculating the area of a trapezoid.
Here I am asking you to derive this result in a more rigourous way: First use Lagrange’s formula and
then integrate the polynomial that you obtained.
2. While proving that the error in the trapezoidal rule is
O
(
h
2
), we used the fact that:
x
4
x
3
(
x

x
3
)(
x

x
4
)
dx
=

1
6
(
x
3
4

3
x
2
4
x
3
+ 3
x
4
x
2
3

x
3
3
)
=

(
x
4

x
3
)
3
6
Prove it.
3.
Theorem.
Suppose
f
(
x
) is twice continously differentiable, then there exist
ξ
n
∈
[
a, b
] such that
b
a
f
(
x
)
dx

I
n
(
f
) =

(
b

a
)
f
(
ξ
n
)
12
h
2
(1)
Corrolary.
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 Spring '10
 THOMAS
 Trapezoid, dx

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