## Introduction increased with each consecutive section in a

Introduction and RationaleInmy math internal assessment, I wanted to explore the concept of integration. Inthis IA, I focused on definite integrals, and I wished to prove that the area betweenthe curve of  bounded by the lines  and ,and the x-axis, is given by the expression  through the use of limits. Idecided to pick this topic because one day, during math class, our teacher provedthat the area under the  curve bounded by the lines  and ,is given by the expression  through the use of limits. We did this bydividing the area under the curve into small sections1 at equal intervals, andevaluating the area under the curve by allowing the width to tend to zero.After class, he had mentioned that the mathematician Pierre de Fermat hadactually proved that the integral of  is  using limits also, but instead of creatingsections under the graph with equal breadth, he used sections under the graphthat were of unequal lengths that increased with each consecutive section in ageometric sequence. This influenced me to pick this IA topic, for I was quiteexcited to discover how Fermat had done it.

IntroductoryProblemSoas an introductory problem, I have tried to prove, as my teacher did in class,that the areas under the   curve bounded by the lines  and ,is given by the expression  through the use of limits. This time, however,I used internal rectangles of unequal lengths that increased with eachconsecutive section in a geometric sequence.  Figure 1Asshown in the diagram above, we have taken the graph of  and have bounded it between lines  and .I have additionally drawn the aforementioned sections under the graph to createan idea as to what I meant by the sections whose breadths are unequal andincreased. So,let us suppose that the  section between  and  starts when .This tells us that the height of that section is given by .Since we know that the breadths are increasing in a geometric sequence, andsince the first section starts at ,we can say that: – … Wherewe assume that .

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Letus suppose that  is the largest value  can take within the bounds of  and .This would mean that ,which gives us the expression: –  ——————————————à(1) Asthe breadth of each rectangle tends to zero, or when  tends to an infinite value, the area of theinner rectangles (the sections under the curve), should be able to provide uswith the area under the curve. The area of each rectangle is given to be thelength multiplied with the breadth of the rectangle. The length is given by the values for each rectangle, whereas the breadthis given by the difference between a rectangle’s  value subtracted from the consecutive value .Hence the area of each inner rectangle is given by: – —————————-à(2) Inorder to calculate the value for the area of all the inner rectangles, we mustadd up the areas of the individual rectangles, hence accounting for allpossible values of .Hence the total area given by the inner rectangles is: – —à(3)Þ   Þ   Þ   Þ   Þ   Þ     The sum of the first  terms of a finite geometric sequence is givenby: –  Wecan use this on the expression  and simplify it to: Þ    However,in order to find the exact answer for the area under the graph, the breadth ofeach section must tend to zero.

As I’ve mentioned before, the breadth of aninner rectangle can be expressed by .This can also be expressed as: –  Sincethe difference of the two terms tends to zero, this essentially means that theterms must be approaching each other to a point of equivalence. Þ   Þ   ——————————————à(4) Hencethe most accurate value for the area under the graph is given when we calculate as ,because when this occurs, the width of the internal rectangles will tend tozero. This means that the sum of all the internal rectangles between  and  will grow closer to actual area under thecurve between those two values.  However,we know from Equation (1) that ,hence we can replace  with the expression and get: – Þ   Þ   Þ   Þ   Þ    Henceproved that  through the use of limits. Deriving the integral of Afterarriving to this conclusion using only limits, I wondered whether it would bepossible to use the very same method to deduce the integral of ,a more generalized function. I assumed it does, and so I replaced  with  and redid the proof: As already calculated from Equation (2), the area of eachinner rectangle is given by: –  Hence the total area given by the innerrectangles is: – Þ   Þ   Þ   Þ   Þ   Þ   Þ   Þ    Asalso previously calculated from Equation(4), the most accurate value for the area under the curve may be deducewhen the widths of the inner rectangles tend to zero.

This occurs when:  Hence,we come closer to our solution by allowing  to tend closer to 1, as shown in theexpression below:  However,we know from Equation (1) that  Þ   Þ   Þ   Þ   Þ    Henceproved that  through the use of limits.             AlthoughI have come to the proof I had initially set out to do, it has proved to be anunsatisfactory ending to my exploration. I wondered if there was more I coulddo with this result and more I can possibly deduce. I initially asked myself ifI could prove the same result using a different method of proof, such as usingthe principle of mathematical induction. However, this proved to be quitedifficult to do, and I have left it for additional work and review. I soonrealized, however, that finding an expression for the integral  is still quite specific, and I began to wonderif I could use the previous proof to find a generalized expression for theintegral of  where the function  is an arbitrary function.

While doing this,however, I realized that finding such an expression using unequal innerrectangles (and hence geometric series) was quite difficult and possiblyimpractical. This also, I have left for further work. I was, however, able toderive an expression for the integral of any function  using equal inner rectangles (and hence arithmeticseries). Since such an expression night not be generalized to such an extent asto be represented by values , and  as the definite integral for  was, I left it in the form of a limit. Hence,my next step was to take my exploration further by deriving an expression forthe integral of an arbitrary function ,but this time, using AP instead of GP. Deriving the expression for integral of Nowthat we are using AP, we can now take the breadth of the inner rectangles underthe curve to be equal again, because each term differs by a constant.

Similarly, the  value for each consecutive inner rectangle,also differs only by a constant, hence allowing the widths of each to by equal.Supposing that the  section between  and  starts when  and that the inner rectangles begin from ,the range of values for  where m is between 0 and  is: … Where is the associated constant for the breadth ofeach rectangle. Given that  is the maximum possible value of  between the bounds  and ,we can say that:  Sincewe know that the integral of the function between bounds  and  is given by the area of the inner rectanglesas the between the bounds as the breadth of the rectangles tends to zero, wecan mathematically express the integral as:  As  is approaching 0,  approaches .

Therefore, this same limit can be rewritten as: —————à(5) This is the expression we were searchingfor, for the integral of . Beforelong, I realized that I had to test this generalization before coming to aconclusion. While wondering what to do, I realized that just as how I derivedthe expression for the integral of ,was it possible to similarly find the expression for ?It is an expression too complicated to calculate using my initial method, butcould it possibly made better by using my generalized expression instead? I putit to the test. Deriving the integral of Therefore,to check if the above expression holds true, I have now once again definedfunction  as:  Let us therefore find the integral of  usingequation (5): Þ   Þ   Þ   Þ   Þ   Þ   Þ   Þ   Þ    Letus now focus on simplifying the following limit to make it easier to calculatethe above expression:   As  and ,we can say that:  Hence, replacing  with ,the limit becomes:  UsingL’Hôpital’s rule, we get: Þ   Þ   Hence, replacing  with ,the expression now becomes: Þ   Þ   Þ   Þ    Which is, indeed, true9. ConclusionAsI have now found the integral for the expression  using limits, having started from trying to deducethe integral for  using limits, I believe it is time to point outthe flaws in my work and scope for further work. Firstly, I believe I assumed  and  to be non-zero constants in all my proofs. Myproof works only when they are both assumed to be non-zero and they will breakdown when either is assumed to be equal to zero.

I believe, this reduces thelegitimacy of my proof to an extent. Furthermore, I proved that the definiteintegral of any function  can be represented in the form:  Where is an arithmetic sequence bounded between  and .However, I only proved this using a test function .I didn’t prove it as rigorously as I would have liked. I believe this too reducesthe legitimacy of my proof, although I still believe it is accurate. If I hadmore space, I believe I could try to find the expression for the definiteintegral  where  and  can take any value, including zero.Furthermore, I have only found an expression for the definite integral for anyfunction  in the form shown above. However, as I’vemention before,  is an arithmetic sequence where  tends to .

Initially, I wanted to find an expression where  is in the form of a geometric sequence, but Iwasn’t able to derive such an expression. I believe that if further work is done,however, I can find this expression. I would also have liked to prove myexpression for the integral of a function  in terms of limits more rigorously. Asaforementioned, I had only proved it using one test function .

This is obviously insufficient in order to prove that the conjecture holds truefor all possible values on .I also wondered whether I could possibly prove my results using mathematicalinduction rather than a direct proof. This is also falls under further work.1 Bysections, I mean to refer to the inner rectangles under the curve. Innerrectangles are a set of adjacent rectangles of a set width between an intervalunder a curve, and where at least one of the upper points lies on the curve. Thismakes it such that the total area of the inner rectangles approximates to thearea under the entire curve in the said interval.

2 Where  is the expression I have used to represent thetotal area of all the inner rectangles.3 Asf(x) is given to be x2.4 Asf(x) is given to be xn.

5 Using the limit Sum law6 This is because, as , the value for  will grow smaller. As  is a constant, we know that the limit tends tozero.7 Using the limit Product law8  which is 1.9 Math HL Data booklet.

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