2014 11-13

What do students at UW get out of Calculus I?Work for the Assessment Committee

Dr. Michelle Chamberlin, Dr. Nathan Clements

November, 13th, 2014

Agenda

1 Methods and context for the study

2 Students overall performance on the final exam

3 Results for four exam questions

4 Conclusions and next steps

5 Questions and discussion

Michelle

Motivation for the study

Each year we are required to assess the learning of ourundergraduate students in mathematics

Needed assessment activities for the 2013-2014 academic year

Decided to draw upon student work that Nathan hadcollected in Calculus I

Michelle

What is assessment of student learning?

A process that all programs at UW are required to undertake.

Involves an iterative process of:

Posing a question about how well are students learning?Gathering evidence about students learningInterpreting and analyzing such evidenceUsing the results to enhance teaching and learning within theprogramRepeat

We are asked to engage in suchcycles yearly

Michelle

2013-2014 Assessment Question

Question: What understandings of the basic concepts and skillsfrom our calculus sequence do students exhibit on thecomprehensive final exams?

Assesses Objective 1 under Goal 1 of our UndergraduateLearning Goals and Objectives

Goal 1: Students shall demonstrate a solid understandingappropriate for the 1000- and 2000-level mathematics requiredin their majors.

Objective 1: Show proficiency in basic skills and conceptsembedded in their courses

Michelle

Data Collected: Student Work on Final Exam

To target our efforts, we elected to focus on Calculus I

Students take a comprehensive and common final

Decided to analyze students work on selected questions fromthe final exam

Provide a picture of student learning across different sectionsand instructors

In Fall 2013, scanned student work on the final exam andentered students overall scores and scores on each questioninto a spreadsheet

To select specific exam items, we looked for questions with aspread of scores

We selected 4 questions from the final exam to analyze

Michelle

Analysis of Student Work on Exam Questions

For each question:

1 Randomly sampled work from approximately 35 students

2 Articulated the procedures, skills, and understandings thatwere associated with each question

3 Read through students work, coding common approaches,different ways of thinking, and errors

4 Generated frequency information about the differentapproaches, ways of thinking, and errors

5 Prepared an overall conclusion about students learning

Michelle

Overall Performance on Final Exam

Mean 70

St Dev 17.91

Q1 59.5

Median 72.5

Q3 84

Math 2200 Exam 4 – All SectionsFall 2013

Overall 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19Mean 69.99 4.44 8.5 2.56 6.66 6.07 2.69 3.42 3.23 3.02 2.3 2.75 1.83 4.62 4.41 4.15 4.87 0.67 1.28 2.53StDev 17.91 2.3 1.08 1.91 2.08 2.65 2.38 1.92 1.99 1.99 1.06 1.41 1.9 1.77 1.08 1.47 1.69 1.25 1.49 1.09Min 7 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0Q1 59.5 3 8 1 6 4 0 2 2 1 2 2 0 3 4 4 4 0 0 3Median 72.5 5 9 3 8 8 4 5 4 4 3 3 1 6 5 5 6 0 0 3Q3 84 6 9 4 8 8 5 5 5 5 3 3 3 6 5 5 6 0 3 3Max 98 7 9 7 8 8 5 5 5 5 3 5 5 6 5 5 6 3 3 3

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Nathan

Question-by-Question SummaryMath 2200 Exam 4 – All Sections

Fall 2013



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Of the 334 students enrolled, 264 took the exam (79%)

219 passed the class= 65% of all students= 82% of students who took the exam.

Nathan

First Question Analyzed

Question 3 Express the integral below as a limit of Riemann sums.Do not evaluate. ∫ π

0sin(5x)dx

Solution limn→∞

n∑i=1

(sin

5πi

n

)π

n

Michelle

1st Q: Understandings and Procedures –∫ π

0 sin(5x)dx

A definite integral represents the area under the curvebetween two given x−values (a and b).We often use a sum of the areas of rectangles to approximatethe integral.This approximation is more accurate as we use rectangles withsmaller widthsThe value of the integral is found by computing the limit asthese widths go to zero, i.e., as the given segment along thex−axis is broken into more and more (an infinite number) ofsubintervals.The width of each rectangle is given by (b − a)/n where n isthe number of partitions and is represented by ∆x .The height of each rectangle is found by evaluating thefunction at any point within each subinterval (x∗i )

Thus,

∫ b

af (x)dx = lim

n→∞

n∑i=1

f (x∗i )∆x

Michelle

1st Q: Understandings and Procedures (cont.) –∫ π

0 sin(5x)dx

In addition, students need to understand:

How the limit relates to the variable n and how this interfaceswith the variable x

How the summation occurs and relates to the variable i

Limit and summation notation (and function notation)

Michelle

1st Q: Results –∫ π

0 sin(5x)dx

Of the 264 students completing the exam, 54 earned full credit,approximately 20%

Of the 35 analyzed:

All but one student understood that an integral represents thearea under the curve, that we use a sum of the areas ofrectangles to approximate this area, and that this is moreaccurate when we use more rectangles

34% were able to determine the width and height for eachrectangle in the limit

89% struggled with the concept and formality of expressingthis idea as a limit

This struggle seemed compounded by students difficulties withunderstanding limit and summation concepts as well as limit,summation, and function notation

Michelle

1st Q: Common Errors –∫ π

0 sin(5x)dx

Taking the limit as x goes to infinity, as x goes to n, or as xgoes to π

Forgetting to write the limit before the summation notation

Forgetting to multiply by the width of each rectangle

Evaluating the sum from n = 0 to n = π or from i = n toi = π

Michelle

1st Question: Conclusion –∫ π

0 sin(5x)dx

Students appear to be leaving Calculus I with an understanding that:

The integral represents the area under the curve

It is estimated by summing the areas of rectangles, becomingmore accurate with more rectangles

However, they struggle with expressing this idea as a formallimit

Likely compounded by difficulties with limit and summationconcepts as well as associated notation

Michelle

Second Question Analyzed

Question 9 Evaluate

∫ 1

0

x

(x2 + 1)3dx

Solution: Requires using integration by substitution whereu = x2 + 1, and so du = 2x dx .∫ 1

0

x

(x2 + 1)3dx =

3

16

Michelle

2nd Q: Understandings and Procedures –∫ 1

0x

(x2+1)3dx

Procedure Understanding or SkillRecognize the integrand asa composed function

Recall substitution as a technique for integrat-ing a composed function

Identify the associatedcompositions of functions:g(x) = x2 + 1,f (x) = 1/x3,

Being able to decompose a composite function

Selecting and substitutingu = x2 + 1

Skillfully selecting a u-substitution suchthat its derivative occurs in the integral,e.g., du = 2xdx , which implies 1

2du = xdx

Determine the antideriva-tive

Application of antidifferentiation formulas andproperties of the integral; in particular∫un du = 1

n+1un+1 + C , n 6= −1

Using appropriate limits ofintegration

Understand that with the change of variables,the coordinate system upon which we are inte-grating has changed

Apply FTC∫ ba f (x) dx =

F (b)−F (a) where F ′(x) =f (x).

Michelle

2nd Q: Results –∫ 1

0x

(x2+1)3dx

Of the 264 students completing the exam, 34% earned full credit.

Of the 35 analyzed:

85% realized the need for u-substitution

However, 89% of these students incorrectly determined theanti-derivative, often misapplying the anti-differentiationformulas

Most students proceeded to evaluate the (often incorrect)anti-derivative with the correct limits of integration

Michelle

2nd Q: Common Errors –∫ 1

0x

(x2+1)3dx

Failing to realize substitution for du includes the x term(leaving the x term in the numerator after u−substitution)

Failing to take an antiderivative before evaluating

Challenges with determining the antiderivative:

Misapplication of antidifferentiation formulas to numerator anddenominator independentlyPerceiving the antiderivative for u3 involves lnIncorrect use of the Power Rule (subtracting the power by 1rather than adding one for the antiderivative)Forgetting to multiply by the coefficient of − 1

2

Michelle

2nd Q: Conclusion –∫ 1

0x

(x2+1)3dx

Approximately 1/3 of students appear to be leaving with anunderstanding of how to use the Substitution Rule to computea definite integral

Most other students are able to recognize whenu−substitution is warranted, are able to decompose theassociated functions, and can make an appropriateu−substitution

Difficulties arise when students try to determine theantiderivative

Nearly all students appear able to apply the FTC

Michelle

Third Question Analyzed

Question 11 Find the value of the limit. If it does not exist, put“DNE”.

limx→∞

ln( 1x )

Solution: −∞

Mean 2.75

St Dev 1.41

Q1 2

Median 3

Q3 3


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Nathan

3rd Q: Understandings and Procedures – limx→∞

ln(1x )

Using limit arithmetic to identify that this limit is of adeterminate, rather than indeterminate, form

Such limit arithmetic in this case includes analyzing the endbehavior of the function as x →∞Understanding that a limit may still exist even when undefinedat a particular point, e.g., for a limit we are interested in thebehavior of the function near a point, not necessarily exactlyat the given point, e.g., just because ln(0) is undefined doesnot necessarily mean this limit does not exist

Understanding that limx→∞

1x = 0.

Understanding that the limx→0

ln x = −∞.

Knowing when to appropriately use L’Hopitals Rule, which isnot called for in this problem.

Nathan

3rd Q: Results – limx→∞

ln(1x )

Of the 264 students completing the exam, approximately 20% com-pleted this item

Of the 30 analyzed:

50% indicated a clear misunderstanding of limit forms(determinate and indeterminate).

Of those 50%, half concluded that the limit does not existsince the determinate form did not exist as a number.

The other half of the 50% misinterpreted the form as anindeterminate form and applied L’Hopital’s rule.

Five of the thirty got the right answer but provided nojustification for their reasoning.

Many misapplied logarithm laws.

Nathan

3rd Q: Conclusion – limx→∞

ln(1x )

Understanding limit behavior in a symbolic way seems difficultfor many students.

Recognizing that an expression, though undefinedalgebraically, produces a meaningful answer “for large x” is askill that is frequently not gained.

This work is evidence that students are not gaining a “bigpicture” conceptual understanding of limits.

Nathan

Fourth Question Analyzed

Question 12 Find the value of the limit. If it does not exist, put“DNE”.

limx→−∞

x2ex

Solution:

limx→−∞

x2

e−x= lim

x→−∞

2x

−e−x= lim

x→−∞

2

e−x= 0

Mean 1.83

St Dev 1.9

Q1 0

Median 1

Q3 3


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Nathan

4th Q: Understandings and Procedures – limx→−∞

x2ex

Recognizing it is an indeterminate form, but that L’Hopital’sRule cannot be applied in its current form

Skillfully writing the product as a quotient such thatapplication of L’Hopitals Rule leads to a form for which thelimit can be determined

Recognizing the need to and applying lHospitals Rule twice

Nathan

4th Q: Results – limx→−∞

x2ex

Of the 264 students completing the exam, 14% earned full credit.

Of the 30 analyzed:

43% either indicated that the limit was of an indeterminateform or applied l’Hospital’s Rule.However, students then struggled with the indeterminateform; most often failing to express the function as an effectivequotient.Other common difficulties included

incorrectly using limit arithmetic,thinking the limit does not exist,and stating that the limit is zero without providing justificationor work.

Three students did determine the limit correctly using anumerical approach, and one student wrote about thedominant nature of ex over x2 .Most students proceeded to evaluate the (often incorrect)anti-derivative with the correct limits of integration

Nathan

4th Q: Conclusion – limx→−∞

x2ex

Identifying and then evaluating limits of indeterminate formsappear challenging for students.

While some students recognized the indeterminate form, theystill presented difficulties with applying l’Hospital’s Rule,especially in this more challenging case that calls for applyingl’Hospital’s Rule twice.

A few students appeared successful however by approachingthis problem through numerical means and by thinking aboutwhich function dominates as x → −∞.

Students’ work on this problem peaked our interest because itallowed use to aske the question: Just how do students thinkabout indeterminate forms?

Nathan

Overall Conclusions

Recall we intentionally selected four questions whereperformance was mixed.

These four items may appear to indicate that students are notmeeting Objective 1 of our Learning Goals: show proficiencyin basic skills and concepts embedded in the course.

However, students did do well on a significant number ofother items on the exam, meaning Calculus I is meetingObjective 1 in many ways.From these other items, it appears that many students areleaving Calculus I understanding

how to differentiate functions;how to graphically interpret functions, derivatives, andanti-derivatives;how to use rates of change to solve real-world problems;how to use approximating rectangles to estimate the areaunder a curve;and how to evaluate integrals.

Nathan

Possible Future Plans

Develop some study resources including possible exercises,class activities, and/or problems that could be used in futuresemesters of Calculus I for preparing for the final exam or inCalculus II as a review at the beginning of the semester.

Use what we learned about students’ understandings in theseareas to write more conceptually-based exam questions forfuture final exams across all calculus courses.

Consider using small-case interviews to further investigatestudents’ thinking on topics that emerged in this analysis;e.g., how do students think about indeterminate forms whenevaluating a limit?

Nathan

Software

2014 11-13