© Boardworks Ltd 2006 1 of 54 Contents © Boardworks Ltd 2006 1 of 54 Using graphs to solve equations The change-of-sign rule Solving equations by iteration

© Boardworks Ltd 20061 of 54

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Using graphs to solve equations

The change-of-sign rule

Solving equations by iteration

Review of the trapezium rule

Simpson's rule

Examination-style questions

Simpson's rule


Simpson's rule

When we used the trapezium rule we split the area that we were trying to find equal strips and then fitted straight lines to the curve. This led to approximations that often had a large percentage error.

Simpson's rule also works by dividing the area to be found into equal strips, but instead of fitting straight lines to the curve we fit parabolas.

The general form of a parabolic curve is y = ax2 + bx + c.

If we are given the coordinates of any three non-collinear points we can draw a parabola through them.

We can find the equation of this parabola using the three points to give us three equations in the three unknowns, a, b and c.


Defining a parabola using three points


Simpson’s rule

Consider the following parabola passing through the three points P, Q and R with coordinates (–h, y0),(0, y1) and (h, y2).

If we take the equation of the parabola to be y = ax2 + bx + c then we can write the area, A, of the two strips from –h to h as:

x0

y

y0 y1 y2

P Q

RWe can use these points to define the area, A, divided by the ordinates y0, y1 and y2, into two strips of equal width, h.

–h h

2= ( + + )h

hA ax bx c dx

3 21 1

3 2= + +h

hax bx cx

3 2 3 21 1 1 13 2 3 2= ( + + ) ( + )ah bh ch ah bh ch

323= + 2ah ch

A


Simpson’s rule

We can find a and c using the points (–h, y0), (0, y1) and (h, y2) to write three equations:

When y = y0 and x = –h :

y0 = a(–h)2 + b(–h) + c = ah2 – bh + c

y1 = c

y2 = ah2 + bh + c

When y = y1 and x = 0:

When y = y2 and x = h :

Adding the first and last equations together gives:

y0 + y2 = 2ah2 + 2c

y0 + y2 = 2ah2 + 2y1

Since c = y1


Simpson’s rule

So,

We can now use this to write the area, A, in terms of the ordinates y0, y1 and y2:

In general, the area under any quadratic function, q(x), divided into two equal strips from x = a to x = b is given by:

2ah2 = y0 + y2 – 2y1

323= + 2A ah ch

It is best not to isolate a because we actually want an expression for .32

3 ah

0 2 1 113= ( + 2 )+ 2h y y y hy

0 2 1 113= ( + 2 + 6 )h y y y y

0 1 213= ( + 4 + )h y y y

q 0 1 213( ) = ( + 4 + )

b

ax h y y y

where h = .2

b a


Simpson’s rule

This forms the basis of Simpson’s rule where we divide the area under a curve into an even number of strips and fit a parabola to the curve across every pair of strips.The area of each pair of strips is taken to be approximately:

If there are four strips (with 5 ordinates) the area will be:

0 1 213 ( + 4 + )h y y y

0 1 2 2 3 413 ( + 4 + )+ ( + 4 + )h y y y y y y

0 1 2 3 413= ( + 4 + 2 + 4 + )h y y y y y

Adding another pair of strips would give the area as:

0 1 2 3 4 5 613 ( + 4 + 2 + 4 + 2 + 4 + )h y y y y y y y

0 1 3 2 413= ( + 4( + )+ 2 + )h y y y y y

0 1 3 5 2 4 613= ( + 4( + + )+ 2( + )+ )h y y y y y y y


Simpson’s rule

In general, for n strips the approximate area under the curve y = f(x) between the x-axis and x = a and x = b is given by:

Where n is an even number and .

This can be more easily remembered as:

f 0 1 3 1 2 4 213( ) ( + + 4( + +...+ )+ 2( + +...+ ))

b

n n nax dx h y y y y y y y y

13 h (yfirst + ylast + 4(sum of yodds) + 2(sum of yevens))

This is Simpson’s rule.

Don’t forget that Simpson’s rule can only be used with an even number of strips (or an odd number of ordinates).

n

abh


x

y

0

Simpson’s rule

Let’s apply Simpson’s rule to approximate the area under the curve y = e–2x between x = 0, x = 2 and the x-axis.

Let’s use four strips as we did when we approximated this area before.

2

y = e–2x

12

y1

1

y2

32

y3 y4

y0

These five points define two parabolas over each pair of strips.

The coordinates of the five points can be found using a table:

1

210x2= xy e 1e

12

32

2e 3e 4e


Simpson’s rule

Using Simpson’s rule with h = gives:1 2

0

xe dx 4 1 3 211+ + 4( + )+ 2

6e e e e

= 0.493 (to 3 s.f.)

12

When we calculated this area using the same number of strips with the trapezium rule we obtained an area of 0.531 (to 3 s.f.).

This result had a percentage error of 8.15%.

Comparing the area given by Simpson’s rule to the actual area gives us a percentage error of:

0.493 0.491×100%

0.491

= 0.41%

Therefore, Simpson’s rule is much more accurate than the trapezium rule using the same number of strips.


Co

nte

nts


Using graphs to solve equations

The change-of-sign rule

Solving equations by iteration

Review of the trapezium rule

Simpson's rule




Examination-style question 1

a) Use Simpson's rule with five ordinates to find an approximate value for

to 3 decimal places.

b) Use integration by parts to find the exact value of the definite integral given in part a).

c) Give the percentage error of the approximation found in part a) to 2 decimal places.

3 2

1lnx x dx

a) If there are five ordinates there are four strips. The width of each strip is therefore

3 1= =

4h

1

2



Using Simpson's rule:

321x2= lny x x

32

52

0 9 34 2ln 4ln2 25 5

4 2ln 9ln3

3 20 1 2 3 41

ln ( + 4 + 2 + 4 + )3

hx x dx y y y y y

9 3 25 54 2 4 2

1= (0 + 4( ln )+ 2(4ln2)+ 4( ln )+ 9ln3)

6

6.998 (to 3 d= .p.)

b) Let and = lnu x 2=dv

xdx

So and1

=du

dx x

3

=3

xv



Integrating by parts:

c) The percentage error is

33 23 32

1 11

ln = ln3 3

x xx x dx x dx

33

1

= (9ln3 0)9

x

19= 9ln3 3

6.776 (to 3 d= .p.)

6.998 6.776×100% =

6.776

3.28%

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© Boardworks Ltd 2006 1 of 54 Contents © Boardworks Ltd 2006 1 of 54 Using graphs to solve equations The change-of-sign rule Solving equations by iteration