Posts Tagged ‘are’

Trapezium Method for Approximating the Area Under a Curve

The trapezium method allows you to approximate the area under a curve by breaking the curve up into and number of trapeziums whose areas can be easily calculated and then adding these areas up. This can be seen below.

A series of trapeziums can be used to approximate the area under a curve

A series of trapeziums can be used to approximate the area under a curve

When using the trapezium method the widths of the trapeziums used can be different, however it is often easier to calculate the total area (using the formula explained later) if all of the trapeziums are of equal width.

To improve the accuracy of the approximation you can use more trapeziums (the process of integration is simply allowing there to be an infinite number of trapeziums)

The area A of a trapezium with height (width when vertical in the approximation) h, and parallel sides a and b is given by


To calculate the approximation we can let the x-values where each of the sides of the trapeziums touch the x-axis be x0,x1,x2… and the y values where they cut the curve be y0,y1

This means the the width of the first trapezium is x1-x0 which we will let equal d which is the same for all the trapeziums if we let their widths be equal. The parallel sides are of length y0 and y1 so the area A0 is given by

A0 = d(y0+y1)/2

The total area A under the curve is found by adding the areas of all of the trapeziums.

A = d(y0+y1)/2 + d(y1+y2)/2 + d(y2+y3)/2 + d(y3+y4)/2 +…….

However since d is a common factor it can brought outside in a bracket. Also all the y co-ordinates occur twice (in two consecutive trapeziums) apart from y0 and the last y co-ordinate but all the terms are also halved this means that all but the first and last y values should be counted once and the first and last halved so we get the trapezium rule as follows

A = d(y1+y2+y3 + … + yn-1 +(y0+yn)/2)

where there are n trapeziums.


Area And Circumference of a Circle : pi

March 2, 2008 58 comments

This site is now at

This is a basic guide to using pi to find the area and circumference of a circle using pi. And also explores why pi makes our formulae work.

Circle radius and circumference

area =πr2

circumference = 2πr or πd

where r = radius and d=diameter


First lets look at the area of a circle, given by area =πr2. This is simple enough to use, we multiply the radius by itself and then by pi.
Does this make sense?
Well r squared is at least going to be an area but it might be a bit small so we multipy by pi. However this doesn’t explain much until we consider what pi is, the easiest way i find to do this is as follows
If we imagine a square that the circle fits inside perfectly(so it touches all four sides like the one above) r squared would give us one quadrant, so the area of that square is 4 x r2 . Of course the circle’s area is a bit smaller so we need to find the ratio between the areas of the square and circle. If we then times this value by four we have a magic constant to multiply r squared by to find the area of a circle (we times by four because we need the area of 4 quadrants and r squared gives us one).
Now this magic constant is pi (which makes sense being just over 3, meaning the area of the circle is just over 3/4 of the area of the square).
The circumference of a circle is given by 2πr or πd. This seems simple, we just multiply the diameter (2r) by our magic constant pi.
Does this also make sense?
seeing as we only have one r this time so only one length it seems we are just finding a factor to increase the length by to make a different length(the circumference) which makes sense.
Again lets consider the square into which our circle fits perfectly, the perimeter of this square would be 4 time the length of one of the sides.
Now the length of the sides = the diameter so the perimeter is 4d.
Notice again that the value we are trying to find for the square is multiplied by 4, but for a circle were going to need a ratio that is a bit smaller.
So we need to replace the 4, for a square, with another, smaller, number — it seems pi will do the job.
To me when I consider pi I don’t look at it as a magical fundamental constant, but more a magical fundamental constant multiplied by four, because when I consider how these formula work using pi this is how they seem to work.
So this new constant is really the ratio of
area of square to area of circle
perimeter of square to circumference of circle.
and it = pi/4 = 0.785398….
so if you have a value for a square and you want a similar value for the circle you just need to multiply it by this number and you’ll have your answer 🙂
I welcome comments, improvements or errors in this post. Please leave your comments below or email me at

By David Woodford