- Level: International Baccalaureate
- Subject: Maths
- Word count: 1396
Area Under a Curve
Extracts from this document...
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.
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 B S 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
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
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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