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Area Under a Curve

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Introduction

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.

image21.pngimage20.pngimage18.pngimage19.pngimage09.pngimage17.pngimage00.pngimage01.png

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.

image18.pngimage17.png

image22.pngimage03.pngimage09.pngimage04.pngimage02.pngimage07.pngimage05.pngimage06.png

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

AB = ½ (0.2)(3.04 + 3.16)

                                   AB = 0.62

Trapezium C:

...read more.

Middle

With the help of technology, create diagrams showing and increasing number of trapeziums.  For each diagram, find the approximation for the area.  

What do you notice?

Number of

Trapeziums

Area – Manual

Calculations

Riemann Sum Application

Graph

SUM

2

3.375

  Y = x2 + 3

image24.pngimage08.png

                                             Y

      A                                   B

                        N

3.375

  A = 0

  B = 1

  N = 2

  T = 2

5

3.34

  Y = x2 + 3

image22.pngimage08.png

                                             Y

      A                                    B

                        N

3.34

  A = 0

  B = 1

  N =

  T = 2

10

  Y = x2 + 3

image25.png

                                             Yimage08.png

      A                                   B

S

                        N                                            

3.335

  A = 0

  B = 1

  N =

  T = 2

20

  Y = x2 + 3

                                             Yimage26.png

      A                                   B  

SDFSDFimage08.png

                        N

3.33375

  A = 0

  B = 1

  N = 2

  T = 2

  • First I calculated the area manually by hand.  Then I tested these results with the use of technology, which in this case was the Riemann Sum application on a TI-84+ silver edition .  Which I used to create the table above.

Observations:

The value of the sum continues to get more and more precise as the number of trapeziums under the curve increases.

Use the diagram below to find a general expression for the area under the curve of g

...read more.

Conclusion

(x0) + g(xn) + 2[g(xi)]

            2n    

x0 : g(x) = x4 + 16x + 2

g(0) = (0)4 + 16(0) + 2

               = 2

xn : g(x) = x4 + 16x + 2

g(1) = (1)4 + 16(1)

               = 19

x1 : g(x) = x4 + 16x + 2

g(0.25) = (0.25)4 + 16(0.25) + 2

                     = 6.00390625

x2: g(x) = x4 + 16x + 2

g(0.5) = (0.5)4 + 16(0.5) + 2

                  = 10.0625

x3: g(x) = x4 + 16x + 2

g(0.75) = (0.75)4 + 16(0.75) + 2

                     = 14.31640625

 = 1 – 0 (0 + 17 + 2[6.00390625 + 10.0625 + 14.31640625]

         2(4)    

     = 9.861328125

Calculator:

f(x)dx = 10.2

y = x5 + 30

 = xn – x0 (g(x0) + g(xn) + 2[g(xi)]

            2n    

x0 : g(x) = x5 + 30

g(0) = (0)5 + 30

               = 30

xn : g(x) = x5 + 30

g(1) = (1)5 + 30

               = 31

x1: g(x) = x5 + 30

g(0.25) = (0.25)5 + 30

                     = 30.000976563

x2: g(x) = x5 + 30

g(0.5) = (0.5)5 + 30

                  = 30.03125

x3: g(x) = x5 + 30

g(0.75) = (0.75)5 + 30

                     = 30.23730469

 = 1 – 0 (30 + 31 + 2[30.000976563 + 30.03125 + 30.23730469]

         2(4)    

     = 30.19238281

Calculator:

Y=  enter y = x5 + 25    2nd      TRACE      7     ENTER   graph displayed   0     ENTER

 1 S  ENTER S

f(x)dx = 30.166667

y = x + 2

          4

 = xn – x0 (g(x0) + g(xn) + 2[g(xi)]

            2n    

x0 : g(x) = x + 2

                          4

g(0) = (0) + 2

                          4

= ½

xn : g(x) = x + 2

                          4

g(1) = (1) + 2

                           4

= 3/4

x 1: g(x) = x + 2

                          4

g(0.25) = (0.25) + 2

                                     4

= 0.5625

x2 : g(x) = x + 2

                          4

g(0.5) = (0.5) + 2

                                 4

= 0.625

x3 : g(x) = x + 2

                          4

g(0.75) = (0.75) + 2

                                    4

= 0.6875

 = 1 – 0 (1/2 + 3/4 + 2[0.5625 + 0.625 + 0.6875]

         2(4)    

     = 0.625

Calculator:

f(x)dx = 0.625

Conclusion:

         My original calculations from my general statement seemed to be very close to the answers that I later found using a calculator. I do not believe that the shapes of the graphs have an effect on the approximation

...read more.

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