We often need to evaluate the definite integral of a function that has no explicit anti-derivative...

Question

We often need to evaluate the definite integral of a function that has no explicit anti-derivative or whose

anti-derivative

is

not

easy

to

obtain.

The

basic

method

involved

in

approximating

∫

b

a

f
(
x
)

dx

is

called

numerical quadrature.

It uses a sum

n
∑

i
=0

a
i
f
(
x
i
) to approximate

∫

b

a

f
(
x
)

dx
.

We

have

seen

one

such

method

from

Math

5A.

Recall

that

the

definite

integral

is

defined

as

a

limit

of

Riemann sums,

so any Riemann sum could be used as an approximation to the integral:

If we divide [
a,b
]

into

n

subintervals

of

equal

length

∆
x

=

b

−
a

n

,

then

we

have

∫

b

a

f
(
x
)

dx

≈

n
∑

i
=1

f
(
x
∗
i

)∆
x
,

where

x
∗
i

is

any

point in the

i
th subinterval [
x
i
−
1
,x
i
].

If

x
∗
i

is chosen to be the left endpoint of the interval, then

x
∗
i

=

x
i
−
1

and we have

∫

b

a

f
(
x
)

dx

≈
L
n

=

n
∑

i
=1

f
(
x
i
−
1
)∆
x
.

If we choose

x
∗
i

to be the right endpoint, then

x
∗
i

=

x
i

and

we have

∫

b

a

f
(
x
)

dx

≈
R
n

=

n
∑

i
=1

f
(
x
i
)∆
x

(as shown in the figure below).

In this project, we are going to discuss three more approximation rules:

Midpoint Rule, Trapezoidal

Rule and Simpson’s Rule
.

1

Midpoint Rule

If

we

consider

the

case

where

x
∗
i

is

chosen

to

be

the

midpoint

̄
x
i

of

the

subinterval

[
x
i
−
1
,x
i
].

Then

we

have

∫

b

a

f
(
x
)

dx

≈
M
n

= ∆
x
[
f
( ̄
x
1
) +

f
( ̄
x
2
) + +

···
+

f
( ̄
x
n
)]
,

where ∆
x

=

b

−
a

n

and

̄
x
i

=

1

2 (
x
i
−
1

+

x
i
)

As shown in the figure below

The following algorithm uses the Midpoint Rule on

n

subintervals to approximate

I

=

∫

b

a

f
(
x
)

dx
:

•

INPUT

endpoints

a
,

b
; positive integer

n
.

•

OUTPUT

approximate solution

XI

to

I
.

•

Step 1

Set

h

=

b

−
a

n

.

•

Step 2

Set

XI

= 0;

Xbar

=

a

+

h

2

:

– Step 3

For

i

= 1
,
2
,
···
n

do Steps 4 to 6.

– Step 4

Set

XI

=

XI

+

f
(
Xbar
)

– Step 5

Set

i

=

i

+ 1

– Step 6

Set

Xbar

=

Xbar

+

h

•

Step 7

XI

=

XI

∗
h
.

•

Step 8 OUTPUT
(XI) and

STOP
.

2

Trapezoidal Rule

Another approximation is called the Trapezoidal Rule:

∫

b

a

f
(
x
)

dx

≈
T
n

=

∆
x

2

[
f
(
x
0
)+2
f
(
x
1
)+2
f
(
x
2
)+
···
+2
f
(
x
n
−
1
+
f
(
x
n
)]
,

where ∆
x

=

b

−
a

n

and

x
i

=

a
+
i
∆
x

The reason for the name Trapezoidal Rule can be seen from the figure below, which illustrates the case with

f
(
x
)

≥
0 and

n

= 4.

The area of the trapezoid that lies above the

i
th subinterval is

∆
x

(
f
(
x
i
−
1
) +

f
(
x
i
)

2

)

=

∆
x

2

[
f
(
x
i
−
1
) +

f
(
x
i
)]

and if we add the areas of all these trapezoids, we get the right side of the Trapezoid Rule.

The following algorithm uses the Trapezoidal Rule on

n

subintervals to approximate

I

=

∫

b

a

f
(
x
)

dx
:

•

INPUT

endpoints

a
,

b
; positive integer

n
.

•

OUTPUT

approximate solution

XI

to

I
.

•

Step 1

Set

h

=

b

−
a

n

.

•

Step 2

Set

i

= 1;

XI

=

f
(
a
) +

f
(
b
);

Xbar

=

a
:

– Step 3

While

i

≤
n

−
1 do Steps 4 to 6:.

– Step 4

Set

Xbar

=

Xbar

+

h

– Step 5

Set

XI

=

XI

+ 2

∗
f
(
Xbar
)

– Step 6

Set

i

=

i

+ 1

•

Step 7

XI

=

XI

∗
h

∗
0
.
5.

•

Step 8 OUTPUT
(XI) and

STOP
.

3

Simpson’s Rule

The above rules for approximate integration results from using straight line segments to approximate a

curve.

(Left/Right

end

point

and

midpoint

use

line

segment

with

0

slope

while

Trapezoidal

Rule

uses

line

segment

with

slope

f
(
x
i
)

−
f
(
x
i
−
1
)

x
i

−
x
i
−
1

.

The

following

approximation

uses

parabolas

instead

of

straight

line

segments.

As before, we divide [
a,b
] into

n

subintervals of equal length

h

= ∆
x

=

b

−
a

n

, but this time we assume

that

n

is

an

even

positive

integer.

Then

on

each

consecutive

pair

of

intervals

we

approximate

the

curve

y

=

f
(
x
)

≥

0

by

a

parabola

as

shown

below.

If

y
i

=

f
(
x
i
),

then

P
i
(
x
i
,y
i
)

is

the

point

on

the

curve

lying

above

x
i
.

A typical parabola passes through three consecutive points

P
i
,P
i
+1
,

and

P
i
+2
.

Although

we

assume

for

the

case

in

which

f
(
x
)

≥

0,

it

is

a

reasonable

approximation

for

any

continuous

function

f

and

this

method

is

called

Simpson’s

Rule

after

the

English

mathematician

Thomas

Simpson.

Please notice that the pattern of coefficients in the formula below are 1
,
4
,
2
,
4
,
2
,
···

,
4
,
2
,
4
,
1.

∫

b

a

f
()

dx

≈
S
n

=

∆
x

3

[
f
(
x
0
) + 4
f
(
x
1
) + 2
f
(
x
2
) + 4
f
(
x
3
) +

···
+ 2
f
(
x
n
−
2
) + 4
f
(
x
n
−
1
) +

f
(
x
n
)]

where

n

is even and ∆
x

=

b

−
a

n

.

The following algorithm uses the Simpson’s Rule on

n

subintervals to approximate

I

=

∫

b

a

f
(
x
)

dx
:

•

INPUT

endpoints

a
,

b
; even positive integer

n
.

•

OUTPUT

approximate solution

XI

to

I
.

•

Step 1

Set

h

=

b

−
a

n

.

•

Step 2

Set X

odd=

a

+

h
; X

even=X

odd+
h
;

i

= 1

Set

XI

=

f
(
a
) +

f
(
b
);

– Step 3

While

i

≤
n

−
1 do Steps 4 and 5:.

– Step 4

If

i

is odd, then set

XI

=

XI

+ 4

∗
f
(X

odd); X

odd = X

odd + 2

∗
h

else, set

XI

=

XI

+ 2

∗
f
(X

even); X

even = X

even + 2

∗
h
;

– Step 5

Set

i

=

i

+ 1

•

Step 7

XI

=

XI

∗
h

3

.

•

Step 8 OUTPUT
(XI) and

STOP
.

4

Please apply the methods introduced above to the following integrals:

•

∫

2

1

√
x
3

−
1

dx

•

∫

π/
2

0

3
√
1 + cos

x dx

•

∫

3

1

e
1
/x

dx

•

∫

3

1

sin

x

x

dx

•

∫

3

−
1

1

8
πe
−
(
x
−
1)2

8

dx

•

∫

3

1

e
x

x dx

integral-1wlpqiji.pdf

We often need to evaluate the definite integral of a function that has no explicit anti-derivative or whoseanti-derivative is not easy to obtain. The basic method involved in approximating∫ baf(x) dx...

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