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The investigation given asks for the attempt in finding a rule which allows us to approximate the area under a curve (I.e. between the curve and the x-axis) using trapeziums

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Introduction

Introduction

The investigation given asks for the attempt in finding a rule which allows us to approximate the area under a curve (I.e. between the curve and the x-axis) using trapeziums (trapezoids).

The first task indicates us to find the approximation of the area of x=0 to x=1 under the curve of gx=x2+3.

gx=x2+3

First, consider the function of gx=x2+3.

Based on the graph, it shows that the area under the curve from x=0 to x=1 is approximated by the sum of the area of two trapeziums. Hence, in order to find the approximation, the formula to find the area of a trapezium has to be considered:

Trapezium = 12×(a+b)×h

Where a = Length of one side

b = Length of parallel side

h = height of the trapezium.

Since there are two trapeziums, it would be simpler to separate into two individual shapes, where the area will be approximated from is at x=0 and x=1. The sum of the area of both trapeziums should provide the overall area under the curve:

Trapezium A. X from 0 to 0.5

h = 0.5

b= 3.25

a = 3

image00.png

Trapezium B. X from 0.5 to 1.0

h = 0.5

b = 4

a = 3.2

image01.pngTrapezium A. is represented from the first graph above. To calculate its area, we would then use the trapezium area formula:

area=12×3+3.25×0.5

area=12×6.25×0.5

area=1.5625 ≈1.56

Trapezium B is represented from the second graph above. Using the same formula:

area=12×3.2+4×0.5

area=12×7.2×0.5

area=1.8

Therefore, the approximate of the function gx=x2+3 is given by
Total area=1.56+1.8=3.36

The trapezium extends past the curve, indicating that there will be an overestimation of the area.

...read more.

Middle

0.05

3.0625

3.09

0.1538125

7

0.05

3.09

3.1225

0.1553125

8

0.05

3.1225

3.16

0.1570625

9

0.05

3.16

3.2025

0.1590625

10

0.05

3.2025

3.25

0.1613125

11

0.05

3.25

3.3025

0.1638125

12

0.05

3.3025

3.36

0.1665625

13

0.05

3.36

3.4225

0.1695625

14

0.05

3.4225

3.49

0.1728125

15

0.05

3.49

3.5625

0.1763125

16

0.05

3.5625

3.64

0.1800625

17

0.05

3.64

3.7225

0.1840625

18

0.05

3.7225

3.81

0.1883125

19

0.05

3.81

3.9025

0.1928125

20

0.05

3.9025

4

0.1975625

Total Area=0.1500625+0.1503125+0.1508125+0.1515625+0.1525625+0.1538125+0.15538125+0.1570625+0.1613125+0.1638125+0.1665625+0.1695625+0.1728125+0.1763125+0.1800625+0.1840625+0.1883125+0.1928125+0.1975625
Total Area=3.3343125 ≈3.334

According to the integration function of Wolfram Mathematica Online, the actual value of the area under the curve would be x2+3 which equals to12xx2+9, hence, subbing the value of x = 1, the approximated area would be 3.333, re-curring. Evidently, by increasing the amount of trapezoids under the curve of the graph, leads to approximated values that were close to the actual value; in this case, the approximate area for twenty trapezoids were closer to the actual value than the approximate area of eight trapezoids. Additional trapezoids make the approximated area more accurate as the values decreases slowly, although very minimal. Possibly due to the fact that the graph is a positive curve, hence trapezoid does not extend much past the curve.

Furthermore, the distance between each trapezoid can be determined via the formula:

position of base ×∆xn+x value of first base

However, the position of the 1st base will be assumed to be in position 0, while the position of the 2nd base is assumed to be in position 1, etc. From the previous graphs, using the obtained values from the table, this seems to be a consistent pattern.

General Expression from 0≤x≤1, g(x) =x^2+3

Let n = number of trapezoids, and A = the approximation of area under the curve.

Based on the question, the general expression is to be found for the area under the curve of g, from x=0 to x=1.

The general expression is very similar to what was done previously, thus, it is simply a succinct form of the area formulas for n

...read more.

Conclusion

rd function given, where its steep curves shows the highest amount of percentage error – 0.450%. Since the general statement revolves around trapeziums, functions that have uneven curves will also cause fiddiculities. In this case, functions that have a parabola would be the most optical graph shape to accompany the general statement.

Although there statement has its limitations and will not always work, it will thoretically work within its boundaries.

Conclusion

Trapeziums can be used in integral calculus to find an approximate area under a curve of a graph, as long as the functions used alongside the general statement is not beyond its limitations.

The general statement which has been achieved in this assignment is:

A=12h[fa+2i=1n-1fa+ih+fb]  

And h is represented byb-an

The general statement obtained is able to work is due to its ability to deal with a wide variety of different functions. It uses the geometrical shape of a trapezium, which proves its uses better than either a rectange or square to suit and adjust to fit approximately around a shape of a curve. It is useful in this aspect since one would wish to minimize and sort of ‘space’ between the shape and the curve, as errors occur where the ‘spaces’ are at. Also, its methods are very simple, yet efficient, and it also have very few limitations, hence, the statement can be used over a wide range of functions.

In conclusion, the general statement would be a useful formula to used in calculating approximate area under curves, alongside the field of integral calculus.

...read more.

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