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# Area Under a Curve

Extracts from this document...

Introduction

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

AB = ½ (0.2)(3.04 + 3.16)

AB = 0.62

Trapezium C:

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  Y      A                                   B                        N 3.375 A = 0 B = 1 N = 2 T = 2 5 3.34 Y = x2 + 3  Y      A                                    B                        N 3.34 A = 0 B = 1 N = T = 2 10 Y = x2 + 3 Y A                                   BS                        N 3.335 A = 0 B = 1 N = T = 2 20 Y = x2 + 3 Y A                                   B   SDFSDF 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

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

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The same techniques and equations will be used in the calculations. It is predicted that since there are more trapeziums, the predicted area will be closer to the actual area (more precise). However, since there are more trapeziums, the equation for the total area is now: TA= AA + AB

1. ## Area under curve

+An = (g0 + g1) + (g1 + g2) + (g2 + g3) +..................................+ (gn-1 + gn). = ((g1 + g2) + (g2 + g3) + (g3 + g4) + ...........................................+ (gn-1 + gn)).

2. ## Comparing the surface area of different shapes with the same volume

The use of this algebraic expressions to find h in terms of r for the cylinder and H in terms of a for the prism. A) 1000cm3= ?r2 x h h= 1000cm3/ ?r2 Surface Area of Cylinder with imputed h Abpcc = ?r2 + 2?rh Abpcc = ?r2 + 2?r (1000cm3/ ?r2)

1. ## Investigating the Ratios of Areas and Volumes around a Curve

The following table demonstrates the results obtained using this program for several values of n, a, and b. Results which do not correlate with the observed pattern (and cannot be put down to the calculator's internal rounding errors) are shown in bold.

2. ## Investigating ratio of areas and volumes

8 1.93E+04 2411.8 8.0 9 6.45E+04 7162.2 9.0 10 2.13E+05 21302 10.0 ? 39.18 12.47 3.14 e 21.47 7.897 2.72 100 1.61E+50 1.61E+48 100 From these results it is evident that the conjecture holds true for the points between x = e and x = ?.

1. ## The aim of this particular portfolio investigation is to find the area under a ...

In the following graph, the number of trapeziums used to divide the area under the curve was increased to 10 trapeziums. Height of each trapezium will be, h= = 0.2. Approximating the area under the graph using 10 trapeziums can be calculated as follows:- Area = area of first trapezium

2. ## Matrix Binomials ...

24-1 = 23 = 8 = Although it appears that Zn = 2n-1(Z), we must check other values to determine that the equation does stay true. When n = 15 Z15 = (Z4)(Z4)(Z4)(Z3) = = Verifying with Zn = 2n-1(Z), Z15 = 215-1 = 214 = 16384 = When n = 9 Z9 = (Z4)(Z4)(Z) • Over 160,000 pieces
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