Trapezium Rule

Error

There was an error loading the page.

Contributors

Simon Thomas

History

○

Simon Thomas 5 years, 2 months ago

Published this.

○

Simon Thomas 5 years, 2 months ago

Created this.

Name	Status	Author	Last Modified
Trapezium Rule	draft	Simon Thomas	17/02/2020 10:26
Trapezium Rule 2	draft	Simon Thomas	17/02/2020 10:44
Mid-Ordinate Rule	draft	Simon Thomas	17/02/2020 10:50

There are 7 other versions that do you not have access to.

Question statement

Use the trapezium rule with {n+1} ordinates ({n} strips)

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
h	number	0.5
n	integer	4
b	number	4
c	integer	3
x	list	[ 2, 2.5, 3, 3.5, 4 ]
y	list	List of 5 items
answer	number	0.17738
exactanswer	number	0.1761417867

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

b
exactanswer
x

This variable doesn't seem to be used anywhere.

Advice

The trapezium rule states that $\int^b_aydx\approx0.5h[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})]$ where $h=\frac{b-a}{n}$

In our example $a=\var{a}, b=\var{b}$ and $n=\var{n}$ is the number of strips.

Hence $h=\frac{\var{b}-\var{a}}{\var{n}}=\var{h}$

So $\int^\var{b}_\var{a}{\frac{1}{x^2+\var{c}}}dx\approx0.5\times\var{h}[(y_0+y_4)+2(y_1+y_2+y_3)]$

Our values of $x_i$ are given by $a+h \times i=\var{a}+\var{h}\times i$, as we vary $i$ from 0 to $\var{n}$

Thus we obtain $[x_0,...,x_\var{n}] = \var{x}$

Now we can calculate each $y_i$ as $y_i = y(x_i) = {\frac{1}{x_i^2+\var{c}}}$

Thus we obtain $[y_0,...,y_\var{n}] = \var{y}$

Hence $\int^\var{b}_\var{a}{\frac{1}{x^2+\var{c}}}dx\approx0.5\times\var{h}[(\var{y[0]}+\var{y[4]})+2(\var{y[1]}+\var{y[2]}+\var{y[3]})] \approx \var{answer}$

For comparison, the exact integral $\int{\frac{1}{x^2+\var{c}}}dx$ is given by $\frac{tan^{-1}(x/ \sqrt {\var{c}})}{\sqrt {\var{c}}}$. Hence the exact value of $\int^\var{b}_\var{a}{\frac{1}{x^2+\var{c}}}dx$ is $\frac{tan^{-1}(\var{b}/\sqrt {\var{c}})}{\sqrt {\var{c}}}-\frac{tan^{-1}(\var{a}/\sqrt {\var{c}})}{\sqrt {\var{c}}} = \var{precround(exactanswer,5)}$

The trapezium rule approximation to the area under the curve can be seen below. The convexity of the curve explains why in this case the trapezium rule gives an overestimate of the true integral.

{plotf(c,n,a,b)}

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Number entry
Number entry		Hasn't run yet

This question is used in the following exams:

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Number entry

Advanced settings

Loading...

Error

Trapezium Rule

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 2 months ago

Simon Thomas 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Number entry

Advanced settings

Generated value: `integer`