Math Portfolio

Type 1 Task

Investigating areas and volumes

Miras International School

Amrebayeva Nurbala Gr. 12IB


In this portfolio, I’m going to investigate ratios of areas and volumes of power functions, which can be generalized as the following:

Y=xn, where n-is the power, nR

This function will be graphed between two arbitrary parameters x=a and x=b such that a<b (in other words boundaries or limits). In this investigation, I’m going to use method of math induction, integration using power rule, application of integration (areas under the curve and solids of revolution), also some knowledge about power functions.

Investigation Process

Given the power function y=x2, graph of which is parabola. I need to consider the region formed by this function from x=0 to x=1 and the x-axis, let’s label this area B; and the region from y=0 to y=1 and the y-axis area A. This can be shown on the illustration (Fig.1) below:


Figure 1

To plot this, I used Graph 4-3 Software, which I found very convenient.

So the areas formed and illustrated above (Figure 1), need to be considered in order to find a ratio between them, hence area A: area B. To find that, I need to use integration using power rule. Integration, using power rule is the general integration rule, which is:

where n-1

It consists of the following steps:

  1. Calculate area B, which is -0
  2. Calculate area of the quadrate (formed by the x and y boundaries of the function), which is
  3. Calculate area A, which is equal to

area A = (area of the quadrate – area B)

       which is  SA=

  1. Step4. Therefore the ratio of the areas, will be


These steps, I repeated for other power functions of the type y=xn, where n is the power, which is the set of the positive integers between x=0 and x=1. The graphs can be seen on the Figure 2

Figure 2

On the figure above, I wanted to show you four functions, which were y=x2, y=x3, y=x4, and y=x5. The functions are labeled and the legend is shown in the right corner. The results, I want present you on the following table:

Tab 1

From the table (Tab 1) I can see that, with change of the power of the function, the ratio of the areas changes as well. I would like also to say that, when n is a positive integer, the graph will look as the following:

Tab 2

From this observation, I can make a certain conjecture, which is

If the power function of the type y=xn, where n – is the power, nZ+ and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to n: 1.

However, notice that my conjecture works or was tested only for the set of positive integers (nZ+).  Now, I need to test my conjecture for other subsets of the real numbers, such as set of integers Z; set of rational numbers Q; set of positive rational numbers Q+; and set of positive real numbers R and irrational numbers.

So, first of all, I will try subset of the rational numbers, - negative integers (I will not test positive integers, because I already done that previously). So, let’s try different negative values of n – power of the function, and we will see if it works or not. I will not show calculations for that (I’m using same method of power integration rule); I will just right the results and present them on the table below (Tab 3).

Join now!

Tab 3

The table above represents the values of the areas and their ratios (SA: SB). In order to see it clearly, I can show you a chart with functions presented there.

Tab 4

I would like to pay attention, on that I’m ignoring the negative sign, when I’m doing a ratio; therefore I would take it as an absolute value - . Therefore my conjecture for the set of negative integers will be as the following:

 If the power function of the type y=xn, where n – is the power, nZ (where  and has limits such as x=0 and ...

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