- Level: International Baccalaureate
- Subject: Maths
- Word count: 2150
Shady areas; math portfolio type 1
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Introduction
Shady areas
This investigation is carried out in order to find a rule to approximate the area under a curve using trapeziums (trapezoids).
From calculus it is known that by the help of integration, the definite area under a curve could be calculated. In this case a geometrical method will be used in order to approximate the area under a curve.
The function g(x) = x2+3 is considered. The graph of g is shown below. The area under the curve from x = 0 to x = 1 is approximated by the sum of the area of two trapeziums.
In order to approximate the area under this curve, the composite trapezoidal rule is used. The composite trapezoidal rule is a numerical approach for approximating a definite integral.
g(x) dx = Area
x2 + 3 dx = Area
According to composite trapezoidal rule:
Area = ½ (width of strip) [first height + 2(sum of all middle heights) + last height]
Since there are two trapeziums from 0 to 1 in fig 1.1 hence the width of each strip: 1/2 = 0.5
x | g(x) or height |
0 | 02 + 3 = 3 |
1/2 | (1/2)2 + 3 = 3.25 |
1 | 12 + 3 = 4 |
So: Area = ½ (0.5) [3 + 2( 3.25) + 4]
Area = 3.37 area units (a.u.)
Increasing the number of trapeziums to five:
Approximating the area in this case with five trapeziums:
Width of every strip: 1/5 = 0.2
x | g(x) or height |
0 | 3 |
1/5 | 3.04 |
2/5 | 3.16 |
3/5 | 3.36 |
4/5 | 3.64 |
1 | 4 |
Area = ½ (1/5) [3 + 2(3.04 + 3.16 + 3.36 + 3.64) + 4]
Area = 3.34 a.u.
Middle
3.34
1.3
3.3359375
1.4
3.334490741
1.5
3.333984375
It is clearly observed that as the number of trapeziums increase, the uncertainty in approximating the area decreases since increasing number of segments of area under the curve are being taken into account. Thereof the value gets closer and closer to the actual value of the integral which is: 3.333333333 (from integration).
In order to find a general expression for the area under the curve of g(x) = x2 + 3 from x = 0 to x = 1, the following diagram is considered with n trapeziums:
As mentioned earlier, area according to composite trapezoidal rule is:
Area = ½ (width of strips) [first height + 2(sum of all middle heights) + last height]
So the general expression in this case for n trapeziums is:
Area = ½ (x) [g (0) + 2(g (x1) + g (x2) +…………….+ g (xn-1)) + g (1)]
[OBS! Width of every strip is x. g (0) = g (x0) and g (1) = g (xn)
Using the results, the general statement that will estimate the area under any curve y = f (x), from x = a to x = b, using n trapeziums could be developed.
Assuming that the following graph is y = f (x):
Composite trapezoidal rule: Area = ½ (width of strips) [first height + 2(sum of all middle heights) + last height]
So the general statement for approximating the area under any curve according to fig 1.7:
Area = ½ (c) [f (a) + 2( f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]
[OBS! Width of every strip is c. f (a) = f (x) since (a = x) and f (b) = f (x+nc) since (b = x+nc)]
Using the general statement with eight trapeziums the area under the following three curves from x = 1 x = 3 could be found:
- y=
- y= 4x3 – 23x2 + 40x – 18
- y=
Area = ½ (c) [f (a) + 2( f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]
Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]
x values | f(x) |
1 | 0.6299605249 |
1+1/4 = 5/4 | 0.7310044346 |
1 +2( 1/4) = 3/2 | 0.8254818122 |
1 +3( 1/4) = 7/4 | 0.9148264275 |
1 +4( 1/4) = 2 | 1 |
1 +5( 1/4) = 9/4 | 1.081687178 |
1 +6( 1/4) = 5/2 | 1.160397208 |
1+7( 1/4) = 11/4 | 1.236521861 |
1 +8( 1/4) = 3 | 1.310370697 |
Area = ½ (1/4) [0.6299605249 +2 (0.7310044346 + 0.8254818122 + 0.9148264275 + 1 + 1.081687178 + 1.160397208 + 1.236521861) + 1.310370697]
Area = 1.980021133 a.u.
Area = ½ (c) [f (a) + 2(f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]
Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]
x values | f(x) |
1 | 2.846049894 |
1+1/4 = 5/4 | 3.474792043 |
1 +2( 1/4) = 3/2 | 3.837612894 |
1 +3( 1/4) = 7/4 | 4.156356486 |
1 +4( 1/4) = 2 | 4.365641251 |
1 +5( 1/4) = 9/4 | 4.484455912 |
1 +6( 1/4) = 5/2 | 4.534134497 |
1+7( 1/4) = 11/4 | 4.534086944 |
1 +8( 1/4) = 3 | 4.5 |
Area = ½ (1/4) [2.846049894 + 2(3.474792043 + 3.837612894 + 4.156356486 + 4.365641251 + 4.484455912 + 4.534134497 + 4.534086944) + 4.5]
Area = 8.265026244 a.u.
- y = 4x3 – 23x2 + 40x – 18
Area = ½ (c) [f (a) + 2(f (x+c) + f (x+2c) +……………..+ f (x+(nc-1)) + f (b)]
Where: c = 2/8 = 1/4 = 0.25, a = 1 and b= 3 in this case. [a = x and b = (x +nc) from fig 1.7]
x values | f(x) |
1 | 3 |
1+1/4 = 5/4 | 3.875 |
1 +2( 1/4) = 3/2 | 3.75 |
1 +3( 1/4) = 7/4 | 3 |
1 +4( 1/4) = 2 | 2 |
1 +5( 1/4) = 9/4 | 1.125 |
1 +6( 1/4) = 5/2 | 0.75 |
1+7( 1/4) = 11/4 | 1.25 |
1 +8( 1/4) = 3 | 3 |
Conclusion
The accuracy of approximation also depends upon the number of trapeziums used. This is evident in figures (1.1 – 1.7) and the data related to them. As the number of trapeziums increases, the approximate value gets closer and closer to the true value of the area.
It could be concluded that the composite trapezoidal rule is exact for polynomials of degree less than or equal to one.
Math Portfolio SL type 1
Shiba Younus IB 2
Date: 23-04-09
Katedralskolan
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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