Shady Areas

Alex Kluivert

Math Methods II IB        5A

2 December 2008

In this investigation you will attempt to find a rule to approximate the area under a curve (i.e. between the curve and the x-axis) using trapeziums (trapezoids).

Consider the function g(x) = x2 + 3

The diagram below shows the graph of g.  The area under this curve from x = 0 to x = 1 is approximated by the sum of the area of two trapeziums.  

Find this approximation.

A = ½ h(b1 + b2)

Trapezium A:

        b1: g(x) = x2 + 3                                        b2: g(x) = x2 + 3

                    g(0) = (0)2 + 3                                                    g(0.5) = (0.5)2 + 3

                                 = 3                                                           = 3.25

AA = ½ (0.5)(3 + 3.25)

      AA = 1.5625

Trapezium B:

        b1: g(x) = x2 + 3                                        b2: g(x) = x2 + 3

                    g(0.5) = (0.5)2 + 3                                            g(1) = (1)2 + 3

                           = 3.25                                                        = 4

AB = ½ (0.5)(3.25 + 4)

                                      AB = 1.8125

Total Area:

A = AA + AB

A = 1.5625 + 1.8125

A = 3.375

Increase the number of trapeziums to five and find a second approximation for the area.

Trapezium A:

b1 : g(x) = x2 + 3                                        b2 : g(x) = x2 + 3

            g(0) = (0)2 + 3                                                    g(0.2) = (0.2)2 + 3

                = 3                                                           = 3.04

AA = ½ (0.2)(3 + 3.04)

                                   AA = 0.604

Trapezium B:

b1 : g(x) = x2 + 3                                        b2 : g(x) = x2 + 3

            g(0.2) = (0.2)2 + 3                                            g(0.4) = (0.4)2 + 3

                           = 3.04                                                           = 3.16

Join now!

AB = ½ (0.2)(3.04 + 3.16)

                                   AB = 0.62

Trapezium C:

        b1 : g(x) = x2 + 3                                        b2 : g(x) = x2 + 3

                    g(0.4) = (0.4)2 + 3                                            g(0.6) = (0.6)2 + 3

                                    = 3.16                                                           = 3.36

AC = ½ (0.2)(3.16 + 3.36)

                                   AC = 0.652

Trapezium D:

b1 : g(x) = x2 + 3                                        b2 : g(x) = x2 + 3

            g(0.6) = (0.6)2 + 3                                            g(0.8) = (0.8)2 + 3

                ...

This is a preview of the whole essay