# 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

by hiddenlol (student)

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

Trapezium B. X from 0.5 to 1.0

h = 0.5

b = 4

a = 3.2

Trapezium 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. Through the values on the graph, one is able to determine the variables.

The second task assigned to this investigation is to increase the amount of trapeziums from the initial two trapeziums to five trapeziums, and then find a second approximation for the area.

The five trapeziums use an identical method as the previous question. Using the method shown, the areas of each trapezium will be calculated and the values will then be added together to obtain the overall approximation of the area.

area=12×3+3.04×0.2

area=12×6.04 ×0.2

area=0.604

area=12×3.04+3.16×0.2

area=12×6.2×0.2

area=0.62

area=12×3.16+3.36×0.2

area=12×6.52×0.2

area=0.652

area=12×3.64+4×0.2

area=12×7.65×0.2

area=0.765

area=12×3.36+3.64×0.2

area=12×7×0.2

area=0.7

Adding all the areas together, we get the result

Total Area=0.604+0.62+0.652+0.7+0.765
Total Area=3.341

Based on the results, the approximation of each area is actually the results of both (a+b) added together. The area is approaching around the value of 3. However, more calculations with additional trapezoids are needed to determine a more precise value.

Eight Trapezoids

Total Area= 0.375976562+ 0.379882812+0.387695312+0.399414062+0.415039062+0.434570312+0.458007812+0.485351562
Total Area=3.335937496≈3.336

The value shows that the area is approaching towards the value of around 3.3. However, more calculations will reveal a more accurate answer.

Twenty Trapezoids

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 ...