Posts Tagged ‘Area’

For the past week or so, we’ve been focused primarily on differential calculus and becoming comfortable with the rules that have been derived through the limiting process. The derivatives of e^x and ln(x) were both derived but only the anti-derivative of the former reveals itself through that process. Since the primitive of ln(x) remains a mystery, only the first part of the exploration below can be tackled.

In order to facilitate completion of the entire sequence below, I will ask the students to determine the derivative of f(x)=-x+xln(x). The result will be very useful for our purposes here and will also launch us into Integration by Parts.

AreaUnder exp


AreaUnder exp2


AreaUnder exp3


Thanks for reading.

When working with composite functions, u-substitution often simplifies the integration process for us. What is sometimes overlooked, however, is the significant adjustment that occurs to the area beneath the curve.  The image and link below allow students to explore the significance of this procedure.

ChainRule Area Cos

Interactively explore u-substitution.


As seen below, u-substitution is not a viable approach to integration as its leads to division by zero. In the case below, and others like it, trigonometric substitution provides a means for us to move forward and solve many problems. The process below also brings trigonometric identities into play, thereby  exposing students to their significance.

Circle Area Trig Substitution Formula1

Circle Area Trig Substitution Formula2

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As I’ve stated before, circle area would once again be investigated. This iteration brings uniform sub-intervals into play (apologies for all the color but thought it would be helpful for students to distinguish upper and lower bounds).  The link directly below this image opens a new tab from which students can adjust two parameters to explore the partitions; those sliders appear at the right-hand side near the bottom.

Circle Area(2)BLOG

In the image above, there are two rectangles displayed; the length of the blue rectangle is based on the circumference of the upper bound, the length of the red on the lower bound. By reducing the width of each partition (link below), you will see that these two lengths approach a common value.

Interactive exploration of the image directly above.


Circle Proof (Uniform)


For a direct comparison between this version and that using non-uniform partitions, I’ve added the image below.

Fermat Circle

One final note here: Area of the circle using annuli of uniform sub-intervals extends naturally to volumes by cylindrical shells. Walls extend perpendicular to the upper and lower bounds of each annulus; the height of these walls is governed by a function of “x” (single variable calculus). For f(x)=c, the layering effect of cylindrical shells forms a cylindrical solid. For f(x)=ax+c, the solid formed is a cone, if f(x)=ax^2+b, a paraboloid emerges. These are a reasonable conclusions that high school students readily accept. Volumes by slabs follow the same argument. 

Cylindrical Shells

Volume Cone Shells4

I also like to have students set up integrals from both dx & dy perspectives (as seen below).

Paraboloid (shells)


Thanks for reading.

I’ve always liked using the “onion proof” as an easy approach to showing students why the area of a circle is what it is. I also strive to fold in as many previously learned concepts as possible when introducing new ones.  Fermat’s non-uniform partitions provide a very rich approach to deriving areas.  This approach as it applies to the circle is shown directly below.  As you will see, factoring skills that students have learned in Grades 10 and 11 are put to work and out pops the circle’s area; students like this.


The Setup

A line on [0,1] was first drawn, then partitions based on q=0.5 identified; those partitions were then summed and the total approached 1 (as it should). This idea was then connected to the formula for sum of infinite geometric series (which was also quickly derived). Additional iterations with q=0.9 and q=0.99 revealed that the non-uniform partitions would become “more uniform” as “q” approached “1”. This idea was then generalized over the interval [0,r] and the lengths of each sub-interval determined.

Concentric circles were eventually drawn around the lower bound of the main interval and the area of each annulus was reasoned out through discussion. It was concluded that the upper bound of each sub-interval could be used to represent the length of each annulus, with the span of the corresponding partition equal to its width. It was agreed upon that these annuli, when”sliced” and “unrolled”, formed rectangles whose areas were easily determined.
Lengths, widths, and ultimately areas of at least 3 of these rectangles were recorded and then summed. Some simple factoring revealed a “new” expression with recognizable components. After some simplification, “q” was replaced with “1” and out popped a very good estimate for the area of a circle.
This process was then repeated for f(x)=x^2 and f(x)=x^3 since the factoring skills required for those was also in place. The entire process was then generalized and applied to the function f(x)=x^n and from this, our expression F(x) was determined.

Shown below are the circle and f(x)=x^n. The links provided here allow students to adjust the common ratios for each and view the resulting partitions.

Non-Uniform Partition Exploration: Circle

Fermat Circle

If we were to sketch a function of area “A” with argument “q”, we would find that it has a point of discontinuity at (1,  πr^2). We can say that the function has a limit of  πr^2 as “q” approaches 1.


Non-Uniform Partition Exploration: f(x)=x^n


Thanks for looking.



Courant, Richard., John, Fritz (1999).  Introduction to Calculus and Analytics: Classics of Mathematics. New York, NY:  Springer-Verlag Berlin Heidelberg.