15.09 The trapezoidal rule

A quadrilateral with two parallel sides is called a trapezoid. The following diagram shows one possible trapezoid shape.

To find the area of a trapezoid we need the lengths of the two parallel sides and the distance between them. We insert a diagonal to make two triangles and then use the fact that the area of a triangle is $\frac{1}{2}\text{base}\times\text{height}$12​base×height.

Thus, the area of the trapezoid must be $\frac{1}{2}ah+\frac{1}{2}bh$12​ah+12​bh or

$A_{\text{trapezoid}}=\frac{h}{2}(a+b)$Atrapezoid​=h2​(a+b)

 

The area of many shapes can be approximated by dividing them into strips with parallel edges so that each strip is close to a trapezoid. It is convenient to make the strips all of the same width. The following diagram shows a shape that has been approximated by subdivision into three trapezoids.

By adding together the areas of the three trapezoids we find that the area of the shape is approximately $\frac{w}{3}\cdot\frac{a+b}{2}+\frac{w}{3}\cdot\frac{b+c}{2}+\frac{w}{3}\cdot\frac{c+d}{2}$w3​·a+b2​+w3​·b+c2​+w3​·c+d2​ and this simplifies to

$A_{\text{shape}}\approx\frac{w}{6}\left(a+2b+2c+d\right)$Ashape​≈w6​(a+2b+2c+d)

With a little thought it should become clear that if there were $n$n trapezoids of equal widths instead of three, then the formula for the area would become

$A\approx\frac{w}{2n}\left(a_0+2a_1+...+2a_{n-1}+a_n\right)$A≈w2n​(a0​+2a1​+...+2an−1​+an​)

where the $n+1$n+1 vertical sides of the $n$n trapezoids have lengths $a_0,a_1,...,a_n$a0​,a1​,...,an​.

Some writers rearrange this area formula, called the trapezoidal rule, into the following form which requires fewer arithmetic operations when applied to a real case.

$A\approx\frac{w}{n}\left(\frac{a_0+a_n}{2}+a_1+...+a_{n-1}\right)$A≈wn​(a0​+an​2​+a1​+...+an−1​)

For the shape illustrated, using three trapezoids appears to give an underestimate of the true area. It may be that a better approximation would be achieved by dividing the area into four trapezoids, or more.

 

Worked examples

Question 1

A certain piece of land has three sides that are lines and a fourth that is an irregular curve, as in the map shown below. The shape of the land is shown twice: once with two trapezoids inscribed and again with three.

We calculate:

$A_{2\ \text{trapezoids}}\approx\frac{1020}{2}\left(\frac{410+360}{2}+290\right)=344250$A2 trapezoids​≈10202​(410+3602​+290)=344250 m$^2$2

$A_{3\ \text{trapezoids}}\approx\frac{1020}{3}\left(\frac{410+360}{2}+310+450\right)=389300$A3 trapezoids​≈10203​(410+3602​+310+450)=389300 m$^2$2

Which estimate seems closer to the true area? To get a better estimate, a surveyor might divide the land into a larger number of narrower strips.

 

Question 2

In a more abstract setting, it is often useful to be able to calculate the area under a graph of a function between two points on the horizontal axis, as in the diagram below.

The area between the graph and the $x$x-axis between the points $x=2$x=2 and $x=7$x=7 has been divided into five trapezoids. The vertical sides of the trapezoids are the ordinates at $x=2,3,4,5,6,7$x=2,3,4,5,6,7. These are respectively
$g(2)=2.88$g(2)=2.88
$g(3)=6.75$g(3)=6.75
$g(4)=6.47$g(4)=6.47
$g(5)=3.58$g(5)=3.58
$g(6)=2.40$g(6)=2.40
$g(7)=6.03$g(7)=6.03

The trapezoid rule gives the following estimate:

$A\approx\frac{7-2}{5}\left(\frac{2.88+6.03}{2}+6.75+6.47+3.58+2.40\right)=23.655$A≈7−25​(2.88+6.032​+6.75+6.47+3.58+2.40)=23.655

 

Practice questions

Question 3

Question 4

Question 5