Open Education Studies, 2020; 2: 220–227
Research Article
Miriam Dagan*, Pavel Satianov, Mina Teicher Improving Calculus Learning Using a Scientific Calculator
https://doi.org/10.1515/edu-2020-0125 received July 2, 2020; accepted September 28, 2020. made possible with calculators. This includes not only the basic computational tools, but also their potential use in Abstract: This article discusses the use of a scientific enhancing conceptual understanding of mathematical calculator in teaching calculus by using representations concepts and approaches, essential for developing of mathematics notions in different sub-languages research and critical thinking by the student. (analytical, graphical, symbolical, verbal, numerical This is very important nowadays, because academic and computer language). Our long-term experience institutions have been dealing with the failure of students shows that this may have a positive and significant effect in the first year of their studies in general and especially on the enhancement of conceptual understanding of in mathematical courses (Lowe and Cook, 2003; Yorke and mathematical concepts and approaches. This transcends Longden, 2004). With technological advances and a large the basic computational uses, and implies a potential flow of students to STEM professions at universities and for real improvement in the learning success, cognitive colleges, there is a constant decline in the mathematical motivation and problem solving skills of the student. basic knowledge of the novice academic students (Gueudet, We illustrate the steps we have taken towards doing this 2013; Bosch, Fonseca & Gascon, 2004). Undergraduate through some examples. students have difficulties in understanding definitions and various representation of mathematical concepts. It is Keywords: scientific calculator; calculus teaching; because teaching and learning in high school is still based different representations; conceptual understanding; more on algorithmic exercises and memorization, rather cognitive motivation. than on deeper understanding of mathematical language and using its computer applications in solving complex and interesting problems. Using the scientific calculator 1 Introduction may contribute to a reduction of this gap between enough developed algorithmic skills of novice engineering Despite the near half-century of their widespread students and insufficiency of research, creative and presence in schools, colleges and universities, educators critical mental skills, required in our current academic have yet to agree on the best way to use electronic pocket studies. calculators in teaching. For example, some claim that the use of electronic pocket calculators harms mathematics learning, while others are indifferent to their presence and are not fully aware of recent steps in their development. 2 Pedagogical strategy Indeed, many school and university teachers simply Our main pedagogical approach is based on the active ignore the possible benefits of calculator use in teaching use of different representations of mathematics notions mathematics, or other fields of STEM (Kissane and Kemp, and problems through diverse sub-languages (analytical, 2013). In academic studies, relatively few educators pay graphical, symbolical, verbal, numerical and computer enough attention to the broad educational opportunities language) of mathematics (Dagan et al., 2018, 2019b). For convenience, we will designate these basic forms of representations by means of the capital first letters of the *Corresponding author: Miriam Dagan, Bar-Ilan University, appropriate words: E-mail: [email protected] A – Analytical representation Pavel Satianov, Shamoon College of Engineering, Israel V – Verbal representation Mina Teicher, Bar-Ilan University
Open Access. © 2020 Miriam Dagan, Pavel Satianov, Mina Teicher, published by De Gruyter. This work is licensed under the Creative Commons Attribution alone 4.0 License. PLUS sub-language, in the context of this article) we must submit it in CN form: PLUS sub-language, in theImproving context Calculus of this Learningarticle) Usingwe must a Scientific submit Calculator it in CN form: 221 10 “Calculate X ”. 10 x1 “Calculate X ”. G – Graphical representation x1 N – Numerical representation C – Computer representation (Calculator depiction, in the context of this article)
For example, the problem: “Calculate 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10” formulated in numerical form and required numerical answer is NN representation of the problem in this classification. If we want to present it in computer language (in CASIO fx-991ES PLUS sub-language, in the context of this article) we must submit it in CN form: 10 Figure 1. Graphical representation“Calculate of the sum X ”. x1 Verbal representation (VN) of the same problem will be “Calculate the sum of natural numbers from 1 to 10”. Graphical representation (GN) of the same problem will be “Calculate the area of the figure shown on figure 1”. Our hypothesis, based on long-term pedagogic 10 research and teaching experience, is: FigureFigure 1. Graphical 1: Graphical representation representation of of the the sum sum X . x1 “The permanent and systematic use of different sub-language representations, and various transitions among them, in mathematical disciplines). To correct this situation, we a calculus course, in addition to extensive use of scientific calculators and graphic applications, yields greater achievement organized special seminars dedicated to the use of the and a higher level of understanding, while enhancing cognitive fx-99IES PLUS CASIO calculator in teaching of mathematics motivation and creative thinking of students”. Such approach is for engineering students. We treat this calculator as an in accordance with the requirements of 21 century that needs interactive math dictionary and, as regards the calculus profound changes in the teaching of mathematics and other course, as a modern machine for evaluating elementary STEM disciplines for engineering students. functions, and, more generally, as a convenient device for storing and processing numerical information. This approach has changed the attitudes of students and teachers regarding the use of the scientific calculator, not 3 Using calculators in teaching only in the calculus course, but also in other math courses calculus and engineering disciplines. Note, that the skillful use of the calculator not only improves the student’s exam In most study programs, the calculus course plays a results, but also enhances the understanding of the key central role. After many years of experience in teaching ideas of calculus, as well as students’ motivation and calculus, we have found that the calculator presents interest in studying them. opportunities for significantly developing exploratory and critical thinking by the student. In this paper, we will describe some of our findings in this direction. Since most of our students have worked with the fx-99IES 4 Classroom teaching examples PLUS CASIO calculator, which allows for mid-term tests, final exams and high school exams, we have used this Consider an example involving simply the calculator in all our examples. As is the case for many skillful use of the scientific calculator. advanced calculators, the lecturer will play a key role in utilizing the full power of this device that transcends its Students are asked to evaluate the following expression: basic computational functions. 2.018 2.018× 2.019 In a survey of lecturers in our department, several 2.019 + − 2.018 2 2 years ago, we found that most lecturers did not use this 2.018 + 2.019 2.018 + 2.019 calculator in the teaching process and some of them were not even familiar with it and its potential for the teaching Although all students had the fx-991ES PLUS CASIO and learning of the calculus course (as well in others calculator, they did this task in a direct “arithmetic” 222 Miriam Dagan, Pavel Satianov, Mina Teicher
Dear Dr. Beata Socha. After carefully reading the proof version of the paper several mistakes weremanner, found: step by step. This appears to be a routine We ask students to calculate the sum. calculation, requiring attention and numerous keystrokes, 1) Onand p.1 linesno more 6-7 ,at than the end that. of the Only article rarely title, theredid we is no observe letter R –a should be 13 + 23 + 33 + 43 + ... + 253 CALCULATOR and was written CALCULATO. student who might be familiar with the useful algebraic
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We ask students to estimate this time both students31000010000 to calculate 15 and instead of a cumbersome of a cumbersome result, 3 they15 see the nice natural number Ans calculateof this a cumbersome enormous result,X sum2.500500025!X3 After 2.500500025 that,10 w10e ask they15 students see the nice to natural calculate number and instead theoretically and practically, comparing twoof data a cumbersome entry result,result,x1 X 2.50050002510 they see see the the nice nice natural natural number Miriam Dagan x1 x100001 methods. To clarify the task we ask the question:50005000. Th “What50005000is is something . Thnumberis is something that makes 50005000. that3 youmakes think you .thinkThis . is something15 that makes you of 50005000a cumbersome. This is somethingresult, thatX makes 2.500500025 you think. 10 they see the nice natural number is an elementary action with a calculator?” The correct think. x1 A answer is “pressing any calculator key”. How many A 3 500050004.3.. ThWeis invite is something students to find that a formulamakes for you quick think summation. X3 for any natural 4.3. We invite students to find a formula for quick summation xX1 for Aany natural clicks will it take directly calculating the given arithmetic 3 4.3. We invite students to find a formula for quick summationx1 X for any natural number A (AA problem according to our classification). We enter the expression? The answer is 9 × 5 + 12 = 57 (45 clicks to x1 number A (AA problemWe inviteaccording students A to our toclassification) find a formula. We enter for the quick A 3 3 X A enter 9 numbers and 12 clicks for arithmeticnumber operationsexpression A ( AA X intoproblem the calculator according∑ and using to the our“CALC classification)” key start to sequentially. We enter3 the A = A AA 4.3. We invite studentsx1 summation to find xa1 formulafor any for naturalquick summation number X ( for any natural expression X 3 into the calculator and using the “CALC” key start to sequentially including “=”). In addition, how many clicks are required A x1 xcalculate1 the3 sumsproblem and record theaccording results obtained: to our classification). We expression X into the calculator and using theA “CALC” key start to sequentially to input an algebraic expression? The answer is 9 × 2 + 10 3 3 calculatenumber the sumsA x 1and(AA record problemA the1: results 1 1 obtainedaccordingAns 1: to∑ Xour classification). We enter the enter the expression x=1 into the calculator = 29 (18 clicks to enter 9 letters and 10 clicks for arithmetic A 2 : 13 23 9 Ans 3 calculate theA Asums 1:and and 13 record 1using theAns resultsthe 1 “CALC” obtained: key start to sequenti- operations. In fact, after that, you will need to press the 3 A 3 : 13 23 33 36 Ans 6 expression X into3 the3 3 calculator and using the “CALC ” key start to sequentially A 2 :A 1 1: 2 13913 Ans3Ans3 31 key “CALC” and twice the key “=” and then to enter xthe1 allyA calculate4 : 1 2 3 the4 100 sums Ans and 10 record the results 3 3 3 3 3 A 3 : A 1 2:2 1 323 36 93 3 AnsAns3 6 3 numbers 2.018 and 2.019 and to press the keycalculate “=”. Therethe sums obtained: andA record5 : 1 the2 3results4 5 obtained225 :Ans 15 3 3 3 3 3 3 3 3 3 3 are ultimately 13 extra clicks. Therefore, computing inA 4 :A 1A3:2 6 1: 3 13223 4 33 1004365 6AnsAns 441 106 Ans 21 3 A 3 1: 3 13 3 13 3 3 3Ans3 1 the algebraic way will require 42 clicks. One may wonder,A 5 :A 1 4 :2 1 3 2 4 3 5 4 225 100 AnsAns 15 10 3 3 6 3 3 3 3 3 3 3 3 3 3 3 what is the significant advantage of the algebraic way?A 6 :A 152:2 : 1 13 224 3594 6 5Ans441 225 3AnsAns 21 15 It is important to understand the major difference in the A 63: : 1 13 32323 33 3343 3653 63 Ans4416 Ans 21
reliability of these two methods. Firstly, we immediately 3 3 6 3 3 A 4 : 1 2 3 4 100 Ans 10 see whether the algebraic expression is entered correctly, A 5 : 13 23 33 463 53 225 Ans 15 secondly, the possibility of error when entering two numbers is much less than when entering nine numbers, A 6 : 13 23 33 43 53 63 441 Ans 21 thirdly we can use this formula for any other numbers, fourthly, future engineers need to consider these issues We see that for each A the sum is an exact square and we 6 for the development of research thinking. can only guess the squares of which numbers. It is easy to see that they are the squares of the sums of the bases of the cubed numbers:
We see that for each A the sum is an exact square and we can only guess the squares of Dear Dr. Beata Socha. After carefully reading the proof version of the paper several mistakes
which numbers. It is easy were to see found: that they are the squares of the sums of the bases of the We see that for each A the sum is an exact square and we can only guess the squares of cubed numbers: Improving Calculus Learning Using a Scientific Calculator 223 1) On p.1 lines 6-7 ,at the end of the article title, there is no letter R – should be which numbers. It isWe easy see to that see for that each they A are the the sum squares is an exactof the square sums ofand the we bases can ofonly the guess the squares of A 1: 13 CALCULATO1 12 R and was written CALCULATO. cubed numbers: which numbers. It is easy to see that they are the squares of the sums of the bases of the A 2 : 13 2)2On3 p.132 line1 242 2in the right column after sign & should be Gascon, 2004). instead Gascon, cubed numbers: 3 3 3 2 2 A 1: 13 1 12 A 3 : 1 .(20042 3 6 1 2 3 . 3 3 2 A 4 : 2 13 23 33 43 102 1 2 3 42 A 2 : 1 2 3 A11: 2 13 1 3)On12 p.3 lines 9-10 (left column) in the formula should be AB instead B 3 3 3 3 3 2 2 3 3 3 A2 5 : 1 2 2 3 4 5 15 1 2 3 4 5 N A 3 : 1 2 3 A 6 2: 13223 3 32 1 22 Figure 2. Graphical representation of the sum f (k) 3 3 3 3 3 3 2 . 2 N A 6 : 1 2 3 4 25 6 21 B 1 2A 3 4 5AB 6 3 3 3 3 32 The3 correct3 2 formula is 2A + − k1 A 4 : 1 2 3 A 4 3: 10 1 21 32 36 4 1 2 3 2 2 Figure 2. Graphical representation of the sum f (k) A + B A + B k1 3 3 3 3 3 2 A(A 12) N . A 5 : 1 U2sing the3 A formula4 4: 5 1132152333Figure413...24 32.A 3Graphical10 42 51 re,presentation2 known 3 4from2 of high the sumschool , wef (k obtain) 4)On p.4 line 12 (left2The column) sum needofA the(A to+ areas 1remove) of thesebrackets rectangles in the beginning is more ofthan equality the area under the curve y = f(x) on 3 3 3 3 3 3 2 + + + + + =2 The sumk 1 of the areas of these rectangles is more than the area under the curve y = f(x) on A 6 : 1 2 3 4Using 5 3the 6 formula3 213 13122333 442 ...5 A6 , known2 kN N 1 A 5 : 1 2 3 4 5 15 1 2 3 224 52 2 kN N 1 from3 3 high Theschool,3 correct we obtainwritten equationthe2 segment A(A should1) [1, be N+1].A (A i.e.1) the following inequality holds: f (k) f (x)dx The1 sum2 3 of... the3A areas 3 1 of these32 3rectangles... A3 is2 more than the areathe under segment the 2 [1,curve N+1]. y = if.e.(x) theon following inequality holds: f (k) f (x)dx A 6 : 1A(A21)3 4 5 6 21 1 2 3 4 5 6 k1 1 2 kN 4 N 1 k1 1 Using the formula 1 2 3 4 ... A , known from high school, we obtain 2 2 2 the segment2 [1, N+1]. i.e. the following inequality holds2 : A(Af (k+) 1) f (x)dxA (A +1) N 3 3 3 On the other hand, theOn sumthe other of the hand, excess the sum areas of the of excessN rectangles areas of Nabove rectangles the curveabove they = curve f(x) y = f(x) Now we can even compete 1 +with2 the+ ... Sigma+ AA Operator(=A(11+)2 (+ ...) of+ theA) CASIO= k1 calculator1Figure and= 2: Graphical representation of the sum ∑ f (k) . Using the formula 1 2 3 4 ... A , known from high school, we obtain k =1 2 2 2 2N 1 4 leave it far behind in calculating2 A(A the sum1): 2 A N(A1 1) 3 3 3 On the other hand, the sum of the excess areas of N rectangles above the curve y = f(x) 1 2 ... A 1 2 ... A i.e. f (x)dx , is thei.e sum. fof(x )areasdx , is ofthe N sum “curved of areas triangles of N “curved” in trianglesfigure 3”, inin figure that all3, inof that them all of them N 1 10 000 2 4 2 5)On p.63 line 4 (left column)2 should be2 k 2instead1 2 N 3 3 3 X 5 00010 0012 150A(A 005 0001) A (A1) 1 i.e2. ...f (x)dxA , is the1 sum 2 of areas... A of N “curved triangles ” in figure 3, in that all of them x1 can be moved into the first rectangle (the highest of all) which has an area equal to f(1). Now we can even compete with the Sigma1 Operator () of the CASIO calculator2 and 4 can be moved into the first rectangle (the highest of all) which has an area equal to f(1). Now we1000000 can even compete with the Sigma Operator (S) of can be moved into3 the first rectangle (the highest of all) which has an area equal to f(1). leave it far behind inNow calculatingAlternatively, we can eventhethe sum even compete CASIO: WeXcalculator with alsoand the somarked Sigmaon and. 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Let f (x) be a decreasing5. Approximate and continuous calculation function of sumsin the using domain integralsx 1, then from figure 2 FigureFigure 3. 3: Graphical Graphical re representationpresentation of of the the differences differences f (k) f (x)dx k N kN N 1 k1 1 Figure 3. Graphical representation of the differences f (k) f (x)dx it is easy to understand Let f that(x) be a fdecreasing(k) is equal and to continuous the sum offunction the areas in theof domainN rectanglesx 1, then from figure. 2 N 1 kN kN N 1 5 Approximate calculation of sumsk1 1 k 1 k N Figure 3. GraphicalTherefore, re thepresentation following inequality of the differences holds: f (k) f (kf)(x)dx f (fx(1))dx it is easy to understand that is equal to the sum of the areas of N rectangles depicted in figure 2. f (k) kN N 1 k1 1 using integrals k1 1 Therefore, the followingk 1 inequality holds: f (k) f (x)dx f (1) k1 Thus, we arrive at a double inequality: depicted in figure 2. 1 kN N 1 Let f (x) be a decreasing and continuous function in the Thus, we arrive at a double inequalityTherefore,: the following inequality holds: f (k) f (x)dx f (1) domain x ≥ 1, then from figure 2 it is easy to understand k1 k =N 7 1 that ∑ f (k) is equal to the sum of the areas of N rectangles 8 k =1 Thus, we arrive at a double inequality: depicted in figure 2. Thus, we arrive at a double inequality: N 8 The sum of the areas of these rectangles is more than Figure 2. Graphical representation of the sum f (k) N 1 N 1 the area under the curve yk =1 f(x) on the segment [1, N+1].
i.e. the following inequality holds: f (x)dx f (1) f (2) f (3) ... f (N) f (x)dx f (1) N 8 The sum of the areas of these rectangles is more than the area under the curve y = f(x) Figureon 2. Graphical representation of the sum1 f (k) 1 N 1 k1 N 1 kN N 1 7 As example, for fN(x1) = 1/x , we get: N 1 the segment [1, N+1]. i.e. the following inequality holds: f (k) f (x)Thedx sum off the(x areas)dx of thesef (rectangles1) f is( more2) thanf the(3 area) under... thef curve(N y) = f(x) on f (x)dx f (1) kN fN 1(x)dx f(1) f (2) f (3) ... f (N) f (x)dx f (1) k1 7 1 the segment1 [1, N+1]. i.e. the following inequality holds: f (k) f (x)dxN 1 1 N 1 k1 1 dx 1 1 1 dx ln( N 11) 1 ... 11 ln( N 1) 1 On the other hand, the sum of the excess areas of N rectangles above the curveOn the yother = hand,f(x) the sum of the excess areas of N rectangles above the curve y = f(x) On the other hand, theAs sumexample, of the for excessN f(1x) = areas 1/x , weof Nget : As example, for f(x) = 1/xx , Nwe1 get:2 3 N x N 1 1 1 As example, for f(x) = 1/x , we get: N 1 i.e. f (x)dx , is the sum of areas of N “curved triangles ” in figure 3, in that all of them rectangles above the curve y = f(x) i.e. f (, xis) dxtheN sum1 f (1 of) f (2) f (3) ... f N(N1 ) f (x)dx f (1) 1 i.e. f (x)dx , is the sum of areas of N “curved triangles” in figure 3, in that all1 of them dxSince for1 large1 values1 of Ndx Nthe1 1differences between ln(NN+1)1 and ln(N) is very small and areas of N “curved triangles” in figurecanln( 3,be Nmovedin that 1into) theall first of rectangle them (the 1 highest of all) which... has an area equal to f(1).dx 1 ln( N1 11) 1 1 dx 1 x 2 3 ln( NN 1) x 1 ... 1 ln( N 1) 1 can be moved into the firstAs example, rectangle for (the f( xhighest) = 1/1x of, we all)tends get : to zero, we get very1 simple approximate formula for large values of N: can be moved into the first rectangle (the highest of all) which has an area equal to f(1). 1 x 2 3 N 1 x which has an area equal to f(1). Since for large values of NN 1the differences between ln(NN1+1) and ln(N) is very small and Therefore, the following inequality holds: dx Since1 for1 large values1 of dxN the differences1 1 between1 ln(N+1) and ln(N) is very small and ln( N 1) 1 ... 1 ln(N 1...) 1 ln( N) tends to zero, we get very simplex approximate2 3 formulaN forx large values of N: 1 tends to zero, we get very1 simple 2approximate3 N formula for large values of N: Since for large values of N the differences between ln(N+1) and ln(N) is very small and Denote1 the1 direct 1calculation result, using operator of CASIO, as S N 1 ... ln( N) 1 1 1 2 3 N N 1 1 ... ln( N) tends to zero, we get very simple approximatek Nformula for large values of N: xN Figure 3. Graphical representation of the differences f (k) f (x)dx 21 31 N1 1 k1 1 SN 1 ... Denote the direct calculation result, using operator of CASIO, as S N N 1 Denotek1N 1the direct1 calculation result, 2using3 operatorN x1 Xof CASIO, as S Therefore, the following inequality holds: f (k) f (x)dx f (1) N 1 k1 1 ... ln( N) Denote2 approximate1 3 1 N 1 resultxN as1 A , that in this case is A ln N . Thus, we arrive at a double inequalityS : 1 ... N 1 1 1 N xN 1 N S 1 ... Denote the direct calculation result, using2 3 operatorN x 1ofX CASION , as S 2 N 3 N x1 X kN N 1 8 AN S N We estimate the relative error of approximation by formula: N (x) 100% Figure 3. Graphical representation of the differences Denotef (k) approximatef (x)dx result as AN , that in this case is ANxN ln N . Denote1 approximate1 1 result 1as A , that in this case is A ln N . S N k1 1 N N SN 1 ... 2 3 N X x 1 AN S N x N kN N 1 We estimate the relative error of approximation by formula: (x) 100% A1 S For N=100 the direct calculationN by CASIO calculator of the sum X Nwe getN Therefore, the following inequality holds: f (k) f (x)dx f (1) We estimate the relative error of approximatS N ion by formula: N (x) 100% Denote approximate result as AN , that in this case is AN ln N . x1 k1 1 S N xN Thus, we arrive at a double inequality: S100 5.1873; A100 ln 1001 4.6052 100 11%xN For N=100 the direct calculation by CASIO calculator of the sum AN weS Nget We estimate the relative error of approximation by formula: (x) X 100% 1 For N=100 the direct calculationN byx 1CASIO calculator of the sum X we get S1000 7.4855; A1000 ln 1000S 6.9078 1000 7% N x1 S 5.1873; A ln 100 4S.605210000 9 . 7876 ; A100011%ln 10000 9.2103 10000 5.9% 8 100 100 100 xN S100 5.1873; A100 ln 1001 4.6052 100 11% S20000 10.4807; A1000 ln 20000 9.9035 20000 5.5% For N=100 the directS1000 calculation7.4855; A1000 by CASIOln 1000 calculator 6.9078 of the 1000 sum 7% X we get S1000 7.4855; A1000x1 ln 1000 6.9078 1000 7% S10000 9.7876Note; A1000 that ln it10000 needs 9about.2103 a qu10000arter 5 .of9% hour for CASIO using direct calculation for S10000 9.7876; A1000 ln 10000 9.2103 10000 5.9% S100 5.1873N; A=10020000 ln.100 4.6052 100 11% S20000 10.4807; A1000 ln 20000 9.9035 20000 5.5% S20000 10.4807; A1000 ln 20000 9.9035 20000 5.5% S1000 7.4855; A1000 ln 1000 6.9078 1000 7% Note that it needs about a Geometricquarter of representationhour for CASIO of the using sum directf (1) +calculation f (2) +f (3) for +... + f (N) as the appropriate S 9.7876; NoteA thatln 10000 it needs 9 .2103about a quarter5.9 %of hour for CASIO using direct calculation for N=20000. 10000 number1000 of rectangles is very easy,10000 but we will get a better integral approximation if we N=20000. S20000 10.4807; A1000 ln 20000 9.9035 20000 5.5% Geometric representation of thethink sum about f (1) the + sumf (2) of + fthe (3) appropriate +... + f (N trapezoid) as the appropriate areas. An integral approximation of the Note that it needs about a quGeometricarter of hourrepresentation for CASIO of usingthe sum direct f (1) calculation + f (2) +f (3)for +... + f (N) as the appropriate number of rectangles is very sumeasy, f but(1) we+ fwill (2) get+f a(3) better +... integral+ f (N) approximationwill be better if wewe think about the sum of the N=20000. number of rectangles is very easy, but we will get a better integral approximation if we think about the sum of the appropriateappropriate trapezoid trapezoid areas. areas An (Figure integral 4). approximation of the Geometric representation of thethink sum about f (1) the+ f sum(2) +off (3)the +...appropriate + f (N) trapezoidas the appropriate areas. An integral approximation of the sum f (1) + f (2) +f (3) +... + f (N) will be better if we think about the sum of the number of rectangles is very easy,sum butf (1) we + will f (2) get + fa (3)better +... integral + f (N approximation) will be better if ifwe we think about the sum of the appropriate trapezoid areas (Figure 4). think about the sum of the appropriateappropriate trapezoid trapezoid areas. areas An (Figure integral 4) approximation. of the
sum f (1) + f (2) +f (3) +... + f (N) will be better if we think about 9the sum of the appropriate trapezoid areas (Figure 4).
9 9
9
224 Miriam Dagan, Pavel Satianov, Mina Teicher
N Figure 4: Approximation of the sum ∑ f (k) as sum of areas of N trapezoids. k =1
N 1 N 1 f (x)dx f (1) f (2) f (3) ... f (N) f (x)dx f (1) Since1 for large values of N the differences between1 ln(N+1) easy, but we will get a better integral approximation if we and ln(N) is very small and tends to zero, we get very think about the sum of the appropriate trapezoid areas. As example, for f(x) = 1/x , we get: simple approximate formula for large values of N: An integral approximation of the sum f (1) + f (2) +f (3) N 1 dx 1 1 1 N 1 dx +... + f (N) will be better if we think about the sum of the ln( N 1) 1 ... 1 ln( N 1) 1 N x 12 13 1N x appropriate Figuretrapezoid 4. Aareaspproximation (Figure 4). of the sum f (k) as sum of areas of N trapezoids 1 1+ + +...+ ≈ ln(1 N) N Sum Nof areasN of N trapezoids depicted in the kfigure1 4 2 3 N Figure 4. Approximation of the sum f (k)Nas sum of areas of N trapezoids Since for large values of N the differences between ln(N+1)Figure and 4.ln( ANpproximation) is very small ofand the sum f (k)Nas sum of areas of N trapezoids Figure 4. ApproximationFigure 4. A ofpproximation theequals sum to off ( kthe)kas 1sum sum of fareas(k) as of sum N trapezoids of areas of N trapezoids Figure 4. ApproximationSumk1 ofof areasthek sum1 of N trapezoidsf (k) as sum depicted of areas in theof N figure trapezoids 4 equals to tends to zero, weDenote get very the simple direct approximate calculation formularesult, using for large ∑ operatorvalues of ofN : k1 k1 S N Sum of areas of N trapezoids depictedf (1) inf the(2) figuref (2) 4 equalsf (3) to f (N 1) f (N) f (1) f (N) CASIO, as Sum of Sumareas of ofSum areas N trapezoids of of areas N trapezoids of depicted N trapezoids depicted in the depictedfigure in the 4 figure equalsin the 4figure to equals 4... toequals to f (1) f (2) ... f (N) 1 1 1 f (1) fSum(2) off (areas2) fof(3 )N trapezoidsf (N 1 depicted) 2f (N) in the 2figure 4 equals to 2f (1) f (N) 2 1 ... f (ln(1) Nf) ( 2) f (2) f (3) ... f (N 1) f (N) f (1) f (2) ... f (N) f (1) f (N) f (1) f (2) xf=N(2f )(1)f (f3()2) f (f2()Nf1()3) f (N) f (N 1) f (N) f (1) f (N) f (1) f (N) 2 13 1 N 1 2 1 2... ... 2 f (1) f (f2()1) ...f (2f )(N...) f (N) 2 SN =1+ + +...+ = 2 f (1) f (2) 2 f (2) f (3) ...This2f (numberN 1) fis( Nclose) fto(1 )thef integral(2) ... off (theN)2 functionf (1) f (fN(x)) on the segment [1, N] 2 ∑ 2 2 2 2 ... 2 f (1) f (2) ...2 f (N) 2 2 3 N x=1 X 2 2 2 2 Denote the direct calculation result, using operatorThis number of CASIO is close, as S toN the integral of the function f(x) on the segment [1, N] N This numberThis numberis close isto close the integral to the integralThisof the number function of the is function closef(x) on to the fthe(x )segment integralon the segment [1,of theN] function [1, N] f(x) fon(1 ) f (N) This number is close to the integral of the function1) f f((1x)) onf (the2) segment... f (N [1,) N] f (x)dx Denote approximate result as A , thatx inNThis this number case is is A close= theto the segment integral [1, of N ]the function f(x) onN the segment [1, N] 1 1 N 1 1 N f (1) fN(N) N 2 1 S 1 ... 1) N N f (1) f (2) ... ff((N1)) f f(N(1) f (N) f (x)dx ln N. 2 3 N X 1) f (1) f(2) ... f (N) 2 f (1) f (Nf ()x)dxN x 1 1) f (1) f (2)1) ...f (1f)(Nf)(2) ... f (N) f (f1)(x)dxf 1(N ) f (x)dx Therefore, we2 get the2 approximate1 formula: We estimate the relative error of approximation by 1)1) f (1) f (2) ... f (N) 1 2 f (x)dx Denote approximate result as A , that in this case is A ln N . 2 1 formula: N Therefore,N we get the approximate formula: 1 N Therefore,Therefore, we get thewe approximateget the approximate formula: formula: f (1) f (N) Therefore, we get theTherefore, approximate we getformula: the approximate 2) f (1) formula:f (2) ... f (N) f (x)dx A − S Therefore, weA get Sthe approximate formula:N N N N N N N f (1) f (N) 1 2 We estimate the relative error of δapproximatN (x) = ion by formula:⋅100% N (x) 2) f (1) 100f (%2) ... f (N) f (x)dxN f (1) f (N) f (1N) f (N) S N 2) S Nf (1) f(2) ... f (N) f (x)dx 2 f (1) f (N) 2) f (1) f (2)2)2) ... f (1f)(Nf)(2)f...(x)dxf1(N) f (x)dx f (1) f (N) Note that for the1 function 2 concave2 downward (Figure 4a) 2) f (1) f (2)1 ... f (N) f (x)dx 2 For N=100 the direct calculation by CASIO calculatorxN of 1 2 x=N Note that for the function1 concave downward (Figure 4a1) −1 kN N For N=100 the thedirect sum calculation ∑ X we byget CASIO Note calculator thatNote for ofthatthe the functionfor sum the function concaveX we concave getdownwardNote that downward for (Figure the function (4Figurea) concave4a) downward (Figure 4a) f (1) f (N) x=1 Note that xfor1 the function concave downward (Figure3) 4 a) f (k) f (x)dx Note that for the functionk Nconcave Ndownward (Figure 4a) N f (1) f (N) 2 kN kNN k 1 1 S 5.1873; A ln 100 4.6052 11% 3) f (k)kN f (fx)(1dx)Nf f(N(1)) f (N) 100 100 100 3)3) f (kk)N f (x)dxN f (1) f (N) 3) f (k)k1 3)f (x)dx1f (k) f (x)dx 2 f (1) f (N) k1 2 S 7.4855; A ln 1000 6.9078 7% k1 In3)1 addition,f1(k) for thef2( xfunction)dx concave upward (Figure 4b) 1000 1000 1000 k1 1 2 2 In addition, for the function concave upwardk 1 (Figure1 4b) S10000 9.7876; A1000 ln 10000 9.2103 10000 5.9% In addition, for the function concavek upwardN (FigureN 4b) In addition,In addition, forIn the addition, functionfor the forfunction con thecave function con upwardcave con upward(Figurecave upward (Figure4b) (Figure 4b) 4b) f (1) f (N) N 4) f (k) f (x)dx S20000 10.4807; A1000 ln 20000 9.9035In addition, 20000 5 for.5% the functionkN concave upward (Figure 4b) N f (1) f (N) 2 kN kNN k 1 1 4)4) f (k)kN f (fx)(1dx)Nf f(N(1)) f (N) 4) f (kk)N f (x)dxN f (1) f (N) Note that it needs about a quarter of hour for CASIO using direct4) calculationf (k)k1 4) ffor (x )dx1f(k) f (x)dx 2 f (1) f (N) Note that it needs about a quarter of hour for CASIO using k1 2 k1 As4)1 examplef1(k) for thef2( xsum,)dx 1 2 3 ... N we obtained double inequality: N=20000. k1 1 2 direct calculation for N=20000. As examplek1 for the sum, 1 + 22+ 3 +...+ N we As example for the sum, 1 2 3 ... N 1 we obtained double inequality: Geometric representation of the sum f (1) + f (2) +f (3) +... + f (N) as the appropriate N k N Geometric representationAs example of Asthe example sum for the f (1) sum,for + f the(2) 1+sum,f (3)2 1obtained3 2... double3N... we inequality: obtainedN we obtained double inequality:double inequality: 1 N 2 N 1 As example for the sum, 1 25) 13 ...2 N3 we... obtainedN doublexdx inequality: k N N number of rectangles+... + f is(N very) as theeasy, appropriate but we will number get a better ofAs rectangles exampleintegral isapproximationfor very the sum, if1 Nwe 2 3 ... N k weN obtained double inequality: N N 1 N k N 2 1 N 21 k 1 3 2 6 N k N 5) 1 2 3 ... N 1xdxN 1 N 2 k k N2 N NN 1 N 1 5) 1 2 3 ... N xdxN 1 N k kNN N2 N 1 think about the sum of the appropriate trapezoid5) areas. 1 An2 integral5)3 ... 1approximation N2 3 x... dxof 1theN xdx21k kN1 N N3 k 2 N 2N 6 N 1 2 k 1 3 2 6 5) 1 2 13 ... 1 N 2 xdxk 1 3 k2 N6 N 1 2 k 1 3 2 6 sum f (1) + f (2) +f (3) +... + f (N) will be better if we think about the sum of the 1 2 k 1 3 2 6 appropriate trapezoid areas (Figure 4).
10 10 10 10 10 10 9
N Figure 4. Approximation of the sum f (k) as sum of areas of N trapezoids k1 Sum of areas of N trapezoids depicted in the figure 4 equals to f (1 ) f (2) f (2) f (3) f (N 1) f (N) f (1) f (N) ... f (1) f (2) ... f (N) 2 2 2 2 This number is close to the integral of the function f(x) on the segment [1, N]
f (1) f (N) N 1) f (1) f (2) ... f (N) f (x)dx 2 1 Therefore, we get the approximate formula:
N f (1) f (N) 2) f (1) f (2) ... f (N) f (x)dx 2 N 1 Figure 4. Approximation of the sum f (k) as sum of areas of N trapezoids Note that for the function concave downward (Figure 4a) k1 N Figure 4. Approximation of the sum Sumf (k )ofas areas sum of areasN trapezoids of N trapezoids depicted in the figure 4 equals to kN N f (1) f (N) 3) f (k) f (xk)dx1 f (1) f (2) f (2) f (3) f (N 1) f (N) f (1) f (N) k1 1 2 ... f (1) f (2) ... f (N) Sum of areas of N trapezoids depicted in the figure 24 equals to 2 2 2 In addition,f (1) f ( 2for) thef (2 function) f (3) concavef (N upward1) f ( N(Figure) 4b) f (1) f (N) ... Thisf (1) number f (2) ...is closef (N )to the integral of the function f(x) on the segment [1, N] N Improving Calculus Learning Using a Scientific Calculator 225 2 2 kN 2 f (1) f (N) 2 4) f (k) f (x)dx f (1) f (N) N 1) This number is close to the integralk1 of the1 function f(x) on2 the segment f [1,(1) N]f (2) ... f (N) f (x)dx Table 1: Outcomes of direct and approximate formulas and2 calculation1 time. f (1) f (N) N As example for the sum, 1 2 3 ... N Therefore, we obtained we double get the inequality:approximate formula: 1) f (1) f (2) ... f (N) N f (x)dx 100 1000 10000 20000 2 1 N k NN 1 N 2 N 1 N f (1) f (N) Therefore,5) 1 we2 get the3 approximate... N formula:xdx ∑ kN N2) N f(1) f(2) 671.46... f (N) f (x)dx21 097.45 666 716.46 1 885 688.59 1 2 kk=1 3 2 6 1 2 N Approximatef (1) f (N) 671.5 21 097.50 666 716.50 1 885 688.63 2) f (1) f (2) ... f (N) f (x)dx Note thatFormula for the 5 function concave downward (Figure 4a) 1 2 Direct calculation time kN 4 sec. N f (1)34 sec.f (N) 330 sec. 615 sec. Note that for the function concave downward (Figure 4a) 3) f (k) f (x)dx k1 2 10 1 kN N f (1) f (N) 3) f (k) f (xIn)dx addition,Table 2: for Outcomes the function of questionnaire concave 1.upward (Figure 4b) 1. I don’t use this calculator and I am not familiar with k1 1 2 kN N f (1) f (N) this devise. In addition, for the function concave upward (Figure Knowing4b) the calculator 4) 1 f (k)2 f (x)3dx 4 2. I use this calculator for computation arithmetic k1 1 2 N Number of teachers 14 10 6 5 operations only. kN f (1) f (N) 4) f (k) f (xAs)dx example for the sum, 1 2 3 ... N we obtained double3. inequality:I am familiar with some basic options of this devise. 2 k 1 1 4. I am familiar with most of options of this devise and N 1 N k N 2 use them.N 1 As example for the sum, 1 2 3 ... N we5) obtained 1 2double 3 inequality:... N xdx k N N 1 2 k 1 3 2 6 N 1 N k N 2 N 1 Questionnaire 2. To what extent do you use CASSIO 5) 1 2 3 ... N xdx k N N fx-991ES Plus in teaching calculus? 2 k 1 3 2 6 1 1. I don’t use calculator in the teaching process. 2. I use this calculator to check computation carried out In Table 1, you can see the direct outcomes (by using during the lesson. operator S of CASSIO fx-991ES Plus10 calculator) and 3. I show students algebraic possibilities of this devise, approximate outcomes (by using formula 5) calculation of and how to use them. this sum for some values of N, as well as the calculation 10 4. I constantly demonstrate the great opportunities of time. this devise in calculus studies. Comparison of approximate results (by using formula 5) with direct computation outcomes (by using operator Concerning the second research question, Questionnaire S of CASSIO fx-991ES Plus calculator), shows the great 3 was completed by Building Department students of accuracy of that approximate formula. Shamoon College of Engineering at the beginning of the first semester of their academic studies (two groups, A and B, with 50 students each). 6 Research questions and Questionnaire 3. To what extent are you familiar assessment experiment with the CASSIO fx-991ES Plus calculator? 1. I use this calculator for arithmetic computation only. Our research questions were: 2. I am familiar with some basic modes of this devise. 1. To what extent do lecturers know and use the 3. I use the algebraic possibilities of this devise. calculator in teaching calculus? 4. I am familiar with most of the options of this devise. 2. To what extent do students know this calculator? 3. To what extent use of the calculator helps you in As an example of our assessment experiment of how calculus learning (in the student’s opinion)? advanced calculator knowledge helps in solving computation problems of calculus, we describe the same Concerning the first research question, Questionnaire 15 minutes test, consisting of two questions, two groups 1 was completed by 35 teachers of the mathematical of first year engineering students (50 students in each department of Shamoon College of Engineering, in the group) mentioned above. In one of the groups, (A), the winter semester 2017. teacher paid particular attention to the opportunities of Questionnaire 1. To what extent are you familiar with the calculator in solving calculus computation problems, the CASSIO fx-991ES Plus calculator? while in the other group, (B), the teacher paid no attention ical calculations. For Question 1, the symbolic calculations result is 226 Miriam Dagan, Pavel Satianov, Mina Teicherical calculationsical calculations. For Question. For 1Question, the symbolic 1, the calculationssymbolic calculations result is result is
x ical calculations. xFor Question3 x 1, thex symbolic1 calculations1 result is ′ f ′(3x) = 2 3 lnx 2⋅3 x ⋅ln3⋅ 1 f (x) = 2f ′(x)ln=22⋅3 ln⋅ln2⋅33⋅ ⋅ln3 ⋅ 2 x 2 x Table 3: Outcomes of questionnaire 2. ical calculationsThe students. For Question of the 1, the “calculator symbolicx calculations expert” result1 group is (A) found ical calculations. For Question′ 1, the= symbolic3 calculations⋅ x ⋅ result⋅ is The students of the “calculator expert” groupf ((A)x) found2 theln value2′ 3 of fln′′(3) using the CASIO The studentsThe of students the “calculator ofthe the “ calculatorvalue expert” group ofexpert (A)f”′ group (found3) using(A) the foundvalue theofthe valuef CASIO(3) of using f ( the 3fx-991ES) CAusing2 SIOx the CA PLUSSIO option d d d x d xx xx 1 [ ] d ′23 =d33≈33153.773⋅≈ xx⋅ ⋅ 1 fx-991ESfx-ical991ES fxPLUS-991ES calculations optionPLUS PLUS option. optionFor xQuestion=[ ] [andand[]x ]=obtainedx[ =1][ , obtained]theand and symbolicfobtained obtained′f(x[()x)= calculations ]x2=2[3[][22 lnln]]xx=2 [3result]⋅≈33153 and is ln⋅..the773773ln3 students3 andand⋅ the the studentsthe students′ students of Using the calculator in 1 2 3 4 Thedx studentsdxdx of the “calculatordx expertdxdx” group (A) found the value2 x of f (3) using the CASIO of the “calculator reject” group (B) calculated the value of expression: 2 x teaching of calculus of theof the “calculator “calculatorthe reject reject “calculator” group” group (B) (B) calculate calculate dreject”dd the the value value group ofof expressionexpression (B)d :: calculated 3 x the value of [x ] 1 2 ≈153.773 Thefx-991ES students PLUS of the′ option “calculator= 3 expertx=[ ]⋅ and” xgroup obtained⋅ (A)⋅ found[ the ]valuex=[3] of f ′(3) usingand the the students CASIO 3 f (x)3 2dx ln 2 3 ln3 dx ′ The studentsexpression: of3 the “calculator3 33 3 expert33 1” group1 (A) found the value of f (3) using the CASIO Number of teachers 20 7 6 2 of the “calculator2 ln 22 reject⋅3 ln”⋅ dgroup2ln⋅⋅3⋅ (B)⋅⋅ lncalculate3⋅⋅ d the2 valuedx 3 ofx expression: ical2 calculationsln 2 3. For Questionln3 1, the symbolic calculations≈ result is fx-991ES PLUS optiond [ ]x=[2] and3 obtained d[2 ]x =[3] 153.773 and the students dx 2 3 dx 3 ′ ≈ The studentsfx-991ES of the PLUS “calculator option expert ”[ group]x=[ ](A) and3 found obtained the value of[ 2f (13]x)= [using3] 153 the CA.773SIO and the students For the Question 2, theof symbolic the “calculator calculations reject result” group is: (B)3 calculated the3 value of expression: For the Question 2, the symbolic calculationsdx 2 resultln is: 2⋅3 ⋅lndx3x ⋅ For the Question 2, the symbolicd calculations result isd: x 3 x 1 Table 4: Questionnaire 3 (group A). of the “calculator reject” group (B) calculatef ′(3 xd) the= 2valueln of2 ⋅expression3 ⋅ln3⋅: fx-991ES PLUS option [ ]x=[ ] and obtained [2 ]x=[3]≈1532.7733 and the students 3 4 1 4 4 4 dx 3 3 dx3 3 4 2 x 4For1 the Question11 2, 1the symbolic2 2 calculations2 ln3 2⋅3 result⋅ln is3: ⋅ of the “calculatorx + reject− +1” group−dx 1(B)= calculate x 2=+dln2 3thex value2−+arctan( of expression−x) : ∫ x+ 2 − 2 dx=3 x2 + ln x3− arctan( 2x)31 ′ ∫x x x +The1 students2 + 3dx of the “calculatorx ln expertx arctan(” group (A)x) found the value of f (3) using the CASIO Knowing the calculator 1 2 3 4 2 ∫ 2 x x 1 2 3ln 2⋅3 ⋅ln23⋅ x x +13 3 2 4 2ForFor the theQuestion Question4 2, 3the symbolic 2, the calculations3 symbolicd result1 is 3calculations: 2d x result is: 2 ln12⋅3 1⋅ln3⋅ 2 2 3 3 ≈ Number of students 22 18 9 The students1 The of students “calculator of “calculator expert” groupfx expert-991ES (A”) group foundPLUS+ ( −theA option) foundnumerical the[ numerical value]x==[ ]of and integral value 2obtained+ of using integral − the[ 2using]x the=[3] 153.773 and the students The students of “calculator[ ] expert ” groupx (A) found2 thedxdx numerical2 3 xvalue lnof integralx dxarctan( using xthe) [ ∫] + 4 For the Question[ ]dx [ 2] ,42 the symbolicx xcalculations1 result 3 3is: CASIO fxCASIO-991ES fxoption-991ES ∫[ option] and∫of [theobtained]dx “calculatorand obtained1 reject 1 ” group (B) 2calculated the value of expression 2 : CASIOFor the fx -Question991ES option 2, the symbolic[[] ]dx and xcalculations+ obtained− result isdx: = x 2 + ln x − arctan(x) The students∫[ ]of∫ “calculator expert2 ” group (A) found the numerical value of integral using the x x +1 33 1 4 [4]4 2 [14] 1 1[] 1 3 3 3 2 4 + − ≈ 3 2 ln 2⋅3 ⋅ln43⋅ Table 5: Questionnaire 3 (group B). x [4] x12+ −dx[ ]1dx3.922dx ≈ 3.9222 2 CASIO∫[2 ]fx-991ES∫[2] option1 12 and obtained The students 1of “xcalculator+1x+ +−∫x1 [−expert] x ”+ 2group1dx2 ≈(A=) foundx the +numericalln x − valuearctan(2 of3 integralx) using the x ∫+ − x dx 2=2 dxx +3ln.922x − arctan(x) ∫ ∫[2] 2 x x [ ]xx +1 3 2 x x +1 3 2 The studentsThe of students “calculator ofThe2 “calculator reject students” groupFor reject the (B)” Question groupcalculate of (B)“calculator[ d[2 4calculate] dx,]two the valuesymbolicd stwo and value calculationsthenexpert”s theirand thensubtraction resultgroup their issubtraction2:: (A) :found the CASIO fx-991ES option ∫[ ] and obtained1 1 The students of “calculator reject” group (B) calculatex +d two− values anddx then≈ 3 .their922 subtraction: Knowing the calculator 1 2 3 43 The students of “calculator3 expert∫[2]” group (A) found2 the numerical value of integral using the The students3 ofnumerical “calculator expert value” group3 (Aof) foundintegral thex numericalx +using1 value theof integral CASIO using the fx-991ES 4 2 2 2 2 4 3 4 2 +ln34 2−+arctan(−4) − [ ] 22−+ln[ ]232−[+4arctan(] −21) 1 4 ln 4 arctan( 4) 2 ln 2 1arctan( 1 2) 2 2 Number of students 24 21 5 0 2CASIO 2 fx-991ESTheoption students option of [“calculator]dx and2and [obtained2 reject] dxobtained”x groupx++ −(B)− calculatedxd two≈=3 .value922x s +andln thenx − theirarctan( subtractionx) : 3 3 CASIO+ −fx-991ES ∫ [option3] − 3 ∫∫+[2] and− obtained 22 4 ln 4 arctan(4) ∫[ 2] ln 2 xarctan(x xx ++211) 3 2 2 3 3 3 3 2 [4] 2 The students2 Theof “ calculatorstudents of 1[reject4 “]calculator” 1group expert (B)2 calculate” groupd ( Atwo) found value thes and numerical then their value subtraction of integral: using the 4 + ln 4 − arctan(x + 4−) − 1dx2 ≈+3ln1.9222 − arctan(2) ∫[2] x2 + [ ]− dx ≈ 3.922 Table 6: Questionnaire 4. 3 3 ∫[x] x +13 3 2 2 CASIO fx-991ES2 option 2x [ ]dxx and+1 obtained 2 + − − ∫[ ] 2 + − The students of “calculator4 ln 4rejectarctan(” group 4(B)) calculate 2d twoln value2 s andarctan( then their2) subtraction: 3 3 1 2 3 The students of “calculator reject” group (B)[4] calculate1 d two1 value s and then their subtraction: 3 3 + − ≈ 2 2 x 2 dx 3.922 4 2 + ln34The− arctan( students4) − of 2“calculator2 + ln 2 −∫3arctan([2] reject”2)x xgroup+1 (B) calculated 2 2 Group A 0 18 32 3 4 2two+ ln values4 − arctan( and 3 4then) − their 2 2 subtraction:+ ln 2 −arctan(2) 3 The students of “calculator 3 reject” group (B) calculated two values and then their subtraction: Group B 31 12 7 2 3 2 3 2 + − − 2 + − 4 ln 4 arctan(4) 2 ln 2 arctan(2) 3 3
to the skillful use of the calculator. The initial level was the same for these groups (see tables 4, 5). That takes more time (if the algebraic features of CASIO Question 1. Find the slope of the tangent line to the fx-991ES PLUS is not used) and so often leads to mistakes.
x=3graph of the function f(x) at the point x=3 Concerning the third research question, questionnaire 4 was completed by the students of groups A and B, x x=3 ( xf 2) 3 mentioned above, at the end of the first semester. x Questionnaire 4. To what extent does use of the 4 3 11 Question 2. Calculate the integral: calculator help you in calculus learning? Question 2. Calculate the integral: x 2 dx )( xf 2) xx 1 2 1. I use this calculator for arithmetic computation and 4 13 11 no more. x dx 2. It helps me to check the results and detect errors. Question 2. Calculate the integral: 2 2 xx 1 3. It helps me not only check errors but also to understand the notion of calculus. These were standard questions13 for both groups and with all symbolic calculations students of each group succeeded equally but while in “calculator expert learned group” 7 Conclusions (A) all students did the numerical calculation correctly and quickly, the students in the other group (B) made a Our long-term experiences of use of the scientific calculator lot of errors and many of them did not have enough time in teaching calculus, and our assessment experiment to complete these numerical calculations. For Question 1, results show, that permanent and systematic use of the the symbolic calculations result is scientific calculator yields greater achievement and a ical calculations. For Question 1, the symbolic calculations result is high level of understanding, while enhancing cognitive
x 1 motivation and creative thinking of students. Note, that f ′(x) = 23 ln 2⋅3 x ⋅ln3⋅ 2 x we always observed surprise and admiration of students,
The students of the “calculator expert” group (A) found the value of f ′(3) using the CASIO when they see unexpected possibilities of mathematical
d d 3 x fx-991ES PLUS option [ ]x=[ ] and obtained [2 ]x=[3]≈153.773 and the students dx dx of the “calculator reject” group (B) calculated the value of expression:
3 1 23 ln 2⋅3 3 ⋅ln3⋅ 2 3 For the Question 2, the symbolic calculations result is:
4 4 1 1 2 3 x + − dx = x 2 + ln x − arctan(x) ∫ x x2 +1 3 2 2 The students of “calculator expert” group (A) found the numerical value of integral using the [ ] [ ]dx CASIO fx-991ES option ∫[ ] and obtained
[4] 1 1 + − ≈ x 2 dx 3.922 ∫[2] x x +1
The students of “calculator reject” group (B) calculated two values and then their subtraction:
2 3 2 3 2 + − − 2 + − 4 ln 4 arctan(4) 2 ln 2 arctan(2) 3 3
Improving Calculus Learning Using a Scientific Calculator 227 and engineering thinking in the process of using their scientific calculator. The above examples and numerous others, demonstrate to students the skillful use of the calculator when studying each topic and each concept of the Calculus course. Note that we also encourage engineering students to investigate the calculator as a computation machine. For example, we direct them to think about calculation speed of this device for those or other computation problems, and about the possible algorithms used by it in these calculations. In this way, the calculator has become an integral part of the studies of Calculus and other mathematics courses we teach.
References
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