## Discussion of exam grading in Math 135, section F2, summer 2006

### The semi exam

Problem#1 #2 #3 #4 #5 #6 Total
Max grade 8 9 10 10 6 6 49
Min grade 2 4 0 3 0 0 17
Mean grade 5.76 6.76 5.12 8.88 2.88 2.53 31.94
Median grade 6 7 3 10 0 2 30

 Letterequivalent Range A B+ B C+ C D F [51,55] [46,50] [41,45] [36,40] [31,35] [26,30] [0,25]

Some general background
I tried to penalize arithmetic errors and "small" errors minimally, unless the problem was materially simplified by the errors. Students needed to read "my" problem and do that problem, though, and not one they invented as a substitute. I also could read only what was written on the exam paper, not what students may have intended. I wanted some supporting evidence for most answers -- some idea of how students knew the answer.

Problem 1 (8 points)
Computing f(2) earned 2 points. Computing f´(2) earned 2 points. Putting these together properly in an equation for the tangent line earned the remaining 4 points.

Problem 2 (9 points)
The answer in each part of the problem was worth 1 point. Supporting reasoning would earn 2 more points for that part.
a) In this case, I was gentle, and awarded 3 points for "plugging in" (sorry, substituting and evaluating) even without any further comment.
b) Here some algebra needed to be shown to earn all 3 points.
c) I wanted some explanation. Certainly citation of limh-->0sin(h)/h=1 is relevant, and earned 1 point. But statements which were false or irrelevant did not earn credit.

Problem 3 (10 points)
1 point for correct evaluation of f(x+h); 2 points for using this in [f(x+h)-f(x)]/h; 2 points for some preliminary algebraic treatment (multiplication by srqt(A)+sqrt(B) top and bottom for suitable A and B); 2 more points for further algebraic treatment; 2 points for final algebraic treatment resulting in the h on the bottom disappearing; 1 point for the final answer.

Problem 4 (12 points)
a) The domain could earn 4 points. The correct answer is [-2,1) and (1,2]. Each endpoint, correctly indicated (that is, "[-2" and "1)" and "(1" and "2]") earned 1 point. Similarly, the range could earn 2 points. Omitting 0 from the range lost a point.
b) Each correct answer here earned 1 point. No partial credit was given and no explanation was needed.

Problem 5 (10 points)
a) (6 points) Getting to the correct system of two linear equations in two unknowns (A and B) was worth 4 points. Solving the equations was worth 2 more points.
b) (4 points) Basically, 2 points for the parabolic part of the graph, and 1 point for each line segment. Of course the graphs should be correctly connected. No credit was earned if what was drawn was not the graph of a function.

Problem 6 (6 points)
2 points for the restriction x&le 8 coming from sqrt(8-x); 2 points for the restriction x> 4 coming from sqrt(x-4) on the bottom (throwing out x=4 was worth 1 of these points); presenting the correct answer was worth 2 points. The correct answer alone would earn 2 points. I wanted to read some supporting evidence for the asserted answer.

### The first exam

Problem#1 #2 #3 #4 #5 #6 #7 #8 Total
Max grade 14 14 18 12 8 10 12 10 97
Min grade 2 0 10 2 1 0 0 0 24
Mean grade 10.58 8.95 14.16 8.37 3.95 7.47 5.74 6.00 65.21
Median grade 13 12 14 10 4 7 3 7 68

19 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

 Letterequivalent Range A B+ B C+ C D F [88,100] [82,87] [73,81] [67,72] [60,66] [51,59] [0,50]

Some general background
Again, I tried to penalize arithmetic errors and "small" errors minimally, unless the problem was materially simplified by the errors. Students needed to read "my" problem and do that problem, though, and not one they invented as a substitute. I also could read only what was written on the exam paper, not what students may have intended. I wanted some supporting evidence for most answers -- some idea of how students knew the answer.

Problem 1 (14 points)
a) (10 points) 1 point for writing the definition of f´(x). 2 points for using this f(x) in the definition correctly. 2 points for combining the resulting fraction correctly. 2 points for canceling terms in "the top of the top". 2 points for canceling h's. 1 point for the correct answer. Students who wrote the answer, without any work or with just the quotient rule, earned no points. Please answer the question which is asked.
b) (4 points) 1 point for finding the slope of the line. 1 point for finding a point the line goes through. 2 points for writing the answer.

Problem 2 (14 points)
a) (8 points) 2 points for using data at x=-2 to deduce a condition on A and B. 2 points for using data at x=1 to deduce a condition on A and B. 2 points for work done towards "solving" for A and B. 1 point each for the correct values of A and B.
b) (6 points) 2 points for a correct straight line to the left of x=-2. 2 points for a parabolic arc opening up with a correct vertex in the interval [-2,1]. 2 points for a correct straight line to the right of x=1. A maximum of 3 points will be awarded if the graph drawn is not continuous.

Problem 3 (20 points)
Each part is worth 5 points, and 1 point will be given for the correct answer. The other points are for some supporting evidence in each case.
a) Factor, cancel, plug in.
b) Use sin(little)/little combined with a product.
c) Algebraic treatment such as factor, plug in, cancel.
d) Plug in. Know the value of cos(Pi) (1 point!).

Problem 4 (12 points)
The graph of y=f´(x) requested has some subtle features. Here is how I will grade answers: 1 point for the overall continuity of the function. I will also expect the answer to be "smooth" since I see little evidence of pathological behavior in the graph of y=f(x).
The graph should intersect the horizontal axes three times: once between c and d, once between d and 0, and once somewhere close to e. Each intersection is worth 1 point.
The graph between the first two intersections should be a bump downwards with a negative minimum (2 points). Between the second and third intersection, there should be a positive maximum on a bump (2 points).
To the left of the first intersection there should be a positive maximum (1 point), and then as x-->the left, the graph should be positive and decrease to 0 (1 point).
To the right of the third intersection there should be a negative minimum (1 point), and then as x-->the right, the graph should be negative and increase to 0 (1 point). If the signs of f´(x) are correct in these intervals and nothing horrible is drawn, then, even if the graph may not have the features I';d like, I'll give 2 of the 4 points possible here.

Problem 5 (8 points)
a) (2 points) 1 point for some correct values of K and L, and 1 point for support of this assertion.
b) (3 points) 1 point each for the {positive|negative} answers, and 1 point for support of the assertions.
c) (3 points) 1 point for a citation of the Intermediate Value Theorem, 1 point for a correct use of the word "continuous" in connection with this function and citation, and 1 point for an appropriate interval in the (x) domain variable.

Problem 6 (10 points)
a) (3 points) Use the Quotient Rule and the Chain Rule. 1 point was lost for each error. b) (3 points) Use the Product Rule. 1 point was lost for each error. c) (4 points) Each result was worth 2 points, and some supporting computation needed to be shown. The first part used the Product Rule, and the second part with the Chain Rule.

Problem 7 (12 points)
2 points for computing the derivation of f(x), and then 1 point for writing f´(x)=0. Finding the roots of the resulting equation was worth 3 points. Finding each of the answers, correctly, was worth 2 points each. If some correct answers are given without supporting reasoning, each correct answer will earn 1 point. If answers are given in the wrong order, 1 point will be lost.

Problem 8 (10 points)
a) (5 points) 2 points for a formula connecting H and D (Pythagoras). 2 points for a solution H=sqrt(82-D2) with only the + sign in front of the square root. Since there is considerable algebraic symmetry between D and H, I took off only 1 point for D as function of H. 1 point for the correct domain. I was willing to be "flexible" about the end points of the domain.
b) (5 points) 2 points for a formula connecting and H (the definition of sine). 1 points for H=8sin(). 2 points for the correct domain. I was willing to be "flexible" about the end points of the domain.

### The take-home semi-exam

Problem#1 #2 #3 #4 #5 Total
Max grade 9 9 9 6 4 35
Min grade 1 0 0 1 0 2
Mean grade 6.5 4.94 5.31 1.49 2.75 22.38
Median grade 7 4 5 2.5 3 21.5

16 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. The chief purpose in giving this exam was to allow students ample time to work on graphing problems where some of the computational details would be quite annoying.

Each of the first 4 problems was worth 9 points. I scored 3 points each for information relating to the function, the first derivative, and the second derivative. I scored these points on more "holistically" than is my custom. That is, I looked at all the information I expected could be obtained from that level of reasoning (including, when necessary, other levels, such as the case of range informaion for the function). I gave 0, 1, 2, or 3 points out of 3 for what I read. I did not attempt to guess at what obscure notations or diagrams meant. Effective communication is part of student responsibility. I felt more justified than usual about this because this was a take-home exam, and students had four days in which to work on it. I did look for explicit answers to my questions. So on the FUNC level, I wanted domain, range, limits, and asymptotes. I looked for supporting reasoning, as well, in this case and in all cases. For example, in problem 4, where the formula for the function has a division, some reason shold be presented for the declaration that the domain is "all x". On the FUNC´ level I wanted a computation of the first derivative (hard to go on at all if the derivative is not correct!) and then identification of critical points and types (max/min/neither) and intervals of increase and decrease, with reasoning when neeeded. For the FUNC´´ level, I wanted a correct computation of the second derivative and then the conclusions which could be drawn by considering it. In particular, I wanted intervals where the graph was concave up and down, and the location of inflection points. Diagnosis of inflection points does not coincide with roots of the second derivative. Here is an example useful to note. The function f(x)=x10 has no changes in concavity and no inflection points -- the graph is a very steep "parabola". But even though f´´(x)=90x8 is 0 when x=0, this function has no. So a place where f´´(x)=0 is not necessarily an inflection point!

In problem 5, I gave 2 points to the part a) answers : identification of f(2) and f´(2) (1 point each). I gave 1 point to the correct answer in part b) (> or <?) and 1 point for supporting reasoning. I read what students wrote, and thus the phrase "rate of change" with no further explanation was not sufficient. I was looking for a reference to concavity. The specific values (8 and 17 and 5) were unimportant here. The key fact is the positivity of the second derivative.

Please realize that students were not asked or required to sketch graphs of any functions in any problems. Initial device-drawn graphs were supplied if students could not use their own calculators to see them. The answers supplied should give enough computational details to supply even the most greedy. I created some graphs of the functions in these problems together with some tangent line segements. They are below, with comments. These graphs are not replacements for the answers expected.

 Problem 1 f(x)=(x-1)/e2x Each of these pictures has a carefully chosen "window", mostly gotten by experimentation. Here x is in [.5,3.5] and y is in [-3.2,3.04]. The graphics program always tries to show a sort of filled square, so all of these pictures are somewhat distorted. For example, here the y range is almost twice the x-range. There's a short horizontal line segment in blue representing the tangent line at the only critical point. The horizontal asymptote (the x-axis) is shown in red. There is one inflection point and a line segment representing the line tangent to the curve at that inflection point is drawn in light green. Please notice that the curve is below this line segment to the left of the inflection point and it is above the line segment to the right of the infliection point. Problem 2 f(x)=(x3+1)/x Here the window is (x) [-2,2] by (y) [-6,4] so there is a 1:5 distortion of horizontal to vertical. The blue (critical point) and red (the vertical asymptote) and light green (inflection point) signal the same types of behavior as in the previous graph. Problem 3 f(x)=x+sin(x) Here the window is square, [-10,10] in both x and y. The graph is the curve which, in effect, wiggles around the main diagonal line, y=x. Look: the curve really does strictly increase as you move along it from left to right. And it has infinitely many critical points (two are shown), each of which is also an inflection point. These have horizontal line segments drawn in blue. Additionally, there is an inflection point on the curve between each pair of critical points. Three of these are shown with their light green tangential line segments. In this case, each critical point is also an inflection point and the curve does appear on both sides of both colors of line segments near the point of tangency. Problem 4 f(x)=(2x2+9)/(x2+x+3) This was the graph whose window was most difficult to select. I finally used [-20,20] in x and [2.5,4.0] in y, a distortion of 24:1. I wanted to make the features visible. The result isn't perfect! The horizontal asymptote is in red. There are two critical points, and their horizontal tangent line segments are in blue. The behavior I find most difficult to see in this picture is near each of the three inflection points. Again, there are light green tangential line segments at each inflection point. If you look at them very closely, maybe you can convince yourself that they actually cut through the graph of y=f(x) in each case. In this window, with the "granularity" of the image, the behavior may not be clear. Problem 5 In this problem, only "point" information about f(x) was given, mostly indirectly though the equation of the line. The window here is [1.4,2.6] by [14.5,21.5], so there's a 1:7 distortion. The line segment shown in magenta is part of the given tangent line, y=5x+7. The curve is the simplest example of a function with f(2)=17 and f´(2)=5 and f´´(2)=8: it is f(x)=17+5(x-2)+4(x-2)2. Certainly any function with the point data given would look locally like this, with the only question being the scale of the locality (how close to x=2 should the window be to look as shown). I hope this may help you understand the answer to b).

### The second exam

Problem#1 #2 #3 #4 #5 #6 #7 Total
Max grade 10 9 11 15 14 12 20 84
Min grade 0 1 0 2 1 0 0 14
Mean grade 4.44 3.22 3.48 8.67 10.67 4.34 6.04 45
Median grade 4 3 6.5 10 11.5 6 5 42.5

18 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

 Letterequivalent Range A B+ B C+ C D F [80,100] [75,79] [65,74] [60,64] [50,59] [45,49] [0,44]

Some general background
Again, I tried to penalize arithmetic errors and "small" errors minimally, unless the problem was materially simplified by the errors. Students needed to read "my" problem and do that problem, though, and not one they invented as a substitute. I also could read only what was written on the exam paper, not what students may have intended. I wanted some supporting evidence for most answers -- some idea of how students knew the answer.

Problem 1 (10 points)
"Implied" arithmetic is o.k. in this problem, but unevaluated function values in f(x), f´(x), and f´´(x) are not o.k. 1 point for F(2).
3 points for F´(2): 2 points for the correct use of the Chain Rule, and 1 point for the answer.
6 points for F´´(2): 2 points for the Product Rule, 2 points for the Chain Rule, 1 point for differentiating the power correctly, and 1 point for the answer.

Problem 2 (12 points)
a) (2 points) Some explicit attention to the top and ln earns 1 point, and attention to the bottom earns the other point.
b) (5 points) 1 point for the answer, and 4 points for the process. A use of L'H counts for 2 points, and the eligibility must be specifically acknowledged.
c) (1 point) The correct equation.
d) (3 points) 1 point for the answer, and 2 points for the process.
e) (1 point) The correct equation.

Problem 3 (16 points)
a) (8 points) 2 points for computing f´(x). 1 point each for detecting the critical numbers. 2 points each for describing the nature of the critical point and why.
Again, please realize that points where the first derivative is 0 may not be relative max/min points. Here is an example: f(x)=x3.
b) (8 points) 2 points for computing f´´(x). 1 point each for detecting where the second derivative vanishes. 2 points each for describing why each represents an inflection point.
Again, please realize that points where the second derivative is 0 may not be inflection points. Here is an example: f(x)=x4.

Problem 4 (16 points)
Preparation (7 points) 2 points for constraint equation; 2 points for the objective function; 3 points for using these to translate to a 1 variable calculus problem, including the domain.
Analysis of the calculus problem (7 points) 2 points for finding the derivative of the function to be maximized. 5 points for the total analysis showing that a maximum has been found. One roadmap: 2 points for checking endpoints, 2 points for evaluating the function at the critical numbers, 1 point for putting the information "together".
Stating the answer (2 points) 2 points for reporting the dimensions of the pasture with largest area.

Problem 5 (14 points)
a) (1 point) Plug in the numbers and check.
b) (7 points) From left to right in the original equation: Chain Rule (2 points); derivative of x3 (1 point); Product Rule (2 points). 2 points for getting a formula for dy/dx. The 2 points for solving for dy/dx can be earned even if there is a mistake in differentiation but only if dy/dx appears twice in the student's previous computation and the successive algebraic manipulations are correct. Also, the 2 points can only be earned for "uninstantiated" dy/dx: that is, no substitutions for x or y have been made (the problem does specifically request an answer "in terms of x and y").
c) (4 points) 1 point for realizing that the line must go through (-2,1), 1 point for the slope, and 2 points for a valid equation of the line.
d) (2 points) The line should go through (-2,1) (1 point) and seem to be tangent (not cross the curve at the point of tangency!). The direction should be correct. (1 point)

Problem 6 (12 points)
2 points for giving a formula for A in terms of r and R. 2 points for finding what A is when r=7 and R=10. 4 points for finding a formula for A´ in terms of r, r´, R, and R´. 2 points for computing A´ at the desired time. 2 points for reporting the answers (decreasing and rate).

Problem 7 (20 points)
a) (12 points) 2 points for identifying the x where the maximum value occurs, either graphically or in words. 4 points each for explaining why f(at "that" x)>f(x) to the left and to the right (in the interval [1,3]). 2 points for stating the conclusion in some fashion.
b) (8 points) 2 points for writing the tangent line approximation in this case, linking f(3.04) and f(3)+f´(3)(.04); 2 points for giving the approximate linearized value; 2 points for detecting the sign of f´´(3) in the graph of f´(x); 2 points for using this information to correctly conclude the relationship between the approximation and the true value of f(3.04). It is also possible to give a convincing argument by just discussing f´(x) (it increases as x increases from 3).

### The final exam

Problem#1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 Total
Max grade 19 12 22 8 14 9 10 12 18 12 12 8 13 15 12 175
Min grade 7 3 0 0 0 0 0 0 0 0 1 0 0 2 0 12
Mean grade 14.41 10.12 18.53 3.76 4.29 6.24 5.06 9.35 10.47 5.59 6.53 6.82 4.42 2.47 7.29 125.73
Median grade 15 12 21 3 3 6 9 11 12 4 6 7 11 6 8 123

17 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam, which are similar to those used during the standard semesters:

 Letterequivalent Range A B+ B C+ C D F [175,200] [165,174] [150,164] [135,149] [110,134] [100,109] [0,99]

Some general background
The statement here would be the same as on previous exams. Most students by now were answering "my" questions, and I noticed a general improvement in giving reasons with words when that was requested: thank you.

Problem 1 (20 points)
Each part was worth 5 points. The answer in each part earned 1 point, and 4 points were awarded for some justification.

Problem 2 (12 points)
1 point for the definition, 2 points for using it with this function, 3 points for converting the compound fraction to a simple fraction, 2 points for "expanding" the square, 3 points for cancellations (additive x2's and multiplicative h's) and 1 point for the limit (the final asnwer).

Problem 3 (22 points)
a) (5 points) 2 points off for a misuse of the Chain Rule.
b) (7 points) 2 points off for a misuse of the Chain Rule. 2 points off for a misuse of the Product Rule.
c) (5 points) 2 points off for a misuse of the Quotient Rule.
d) (5 points) An answer alone is fine, but of course I welcomed any mention of the Fundamental Theorem of Calculus.

Problem 4 (8 points)
a) (4 points) 2 points for the correct computation of f´(x). 2 points for a verification of the assertion, which should include use of (-)(-)=+ and the evenness of 300.
b) (4 points) Direct citation of the Mean Value Theorem is welcome, but I also accepted a use of this consequence: if f´(x)>0, then f(x) is increasing. This "theory" (worth 2 points) had to be attached to the function in this problem, and this "attachment" correctly done was worth the last 2 points.

Problem 5
Preparation (8 points) 2 points for constraint equation; 2 points for the objective function; 3 points for using these to translate to a 1 variable calculus problem, including the domain.
Analysis of the calculus problem (7 points) 2 points for finding the derivative of the function to be maximized. 5 points for the total analysis showing that a maximum has been found. One roadmap: 2 points for checking endpoints in some manner, 2 points for evaluating the function at the critical number (just checking that it is postive is sufficient!), 1 point for putting the information "together".
Stating the answer (2 points) 2 points for reporting the dimensions of the box with largest volume.

Problem 6 (10 points)
Horizontal asymptotes: (6 points) Two equations should be the answer, and each correct answer will earn 1. The verification that these answers are correct will each earn 2 points: an appropriate limit must be computed.
Vertical asymptote: (4 points) The answer (one equation) will earn 1 point. Detection that the bottom is 0 at a certain value of x earns 2 points, but this only establishes "eligibility" for a vertical asymptote. Additionally, the student must verify (or even mention!) that the top is not 0 at that value of x, and this will earn 1 point.
The problem statement carefully requests <1>equation of the asymptotes. If equations are not given, 1 point will be deducted from the score.

Problem 7 (10 points)
This was a "simple" problem but the questions were written so briefly that they seemed (in retrospect!) almost designed so students would not answer the questions appropriately. It is certainly true that a function which is not defined at x=a cannot be continuous or differentiable at x=a. Therefore a rather trivial way to answer the question would be to give as example (for both parts!) any function whose domain doesn't include some number. I will give some credit (2 points of 5 in each part) to an answer of that type. I believe such an answer essentially "trivializes" the question and does not tell me if a student knows much about continuity or differentiability.
On the other hand, there are contrasting techniques for attempting "good faith" answers to the questions which were asked. An algebraic definition of a function can be given. Then the student may try to show that the function given does not have some necessary attribute of the definition involved -- that some aspect of the definition is violated. A student could also attempt to answer the question geometrically, by drawing the graph of a function. Much of the course has involved considering and contrasting numerical and algebraic and geometric properties of functions so certainly here (regardless of the intentions of the questioner!) a student can earn full credit with either approach. In each part, 1 point will be given for an example, and 4 points for a discussion or computation validating the example. The discussion or computation must include some reference to a key part of the definition of continuity or the definition of derivative. This must be given explicitly in words or symbolically using appropriate algebra (such as certain limits not being equal).
If I give a question of this type again, I will certainly make the statement more longer and more precise. Then I'd hope to read more suitable answers.

Problem 8 (12 points)
a) (6 points) Differentiation is worth 5 points (differentiation of the terms on the left side of the equation are worth 1 and 2 and 2 points). Solving for dy/dx correctly earns 1 point.
b) (4 points) 1 point for realizing that the line must go through (3,1), 1 point for the slope, and 2 points for a valid equation of the line.
c) (2 points) The line should go through (3,1) (1 point) and seem to be tangent (not cross the curve at the point of tangency!). The direction should be correct. (1 point)

Problem 9 (18 points)
a) (10 points) 3 points for the critical numbers, 3 points for conclusions about the nature of the critical numbers, and 4 points for supporting evidence, such as signs of f´(x).
b) (4 points) 1 point for m and 1 point for M. 1 point for continuity and 1 point for correct {in|de}creasing behavior.
c) (4 points) 1 point for the correct number of inflection points, and 1 point each for correctly marking them on the graph.

Problem 10 (12 points)
2 points for giving a formula for V in terms of r. 4 points for finding a formula for V´ in terms of r and r´. 2 points for computing r at the volume given. 4 points for combining it all into the answer.
Students who don't have an expression connecting V´ and r and r´ are not doing this problem and can earn only the 2 formula points and the 2 points for getting a specific value of r.

Problem 11 (12 points)
a) (8 points) 1 point for the derivtive. 2 points for the critical numbers. 3 points for testing, which must include the endpoints or else just 1 point. 2 points for reporting the correct answers which are athe values of the function. Students who ignore the endpoints will get 4 out of 8 for this part of the problem (1+2+1+0).
b) (4 points) 2 points for the correct answer or for an answer which follows from the student's answer to a). 2 points for mentioning the Intermediate Value Theorem.

Problem 12 (8 points)
Length of the subintervals (2 points), correct sample points (2 points), computation and answer (4 points).
1 point deducted for not matching an interval with its correct sample point. 2 points deducted if there are not 3 subintervals.

Problem 13 (13 points)
1 point will be deducted for the problem score if "+C" is missing from any of the final answers.
a) (3 points) Powers of x and algebra.
b) (5 points) Substitution, antidifferentiation, solution.
c) (5 points) Substitution, antidifferentiation, solution. A bit of algebra.

Problem 14 (15 points)
4 points for the graph, and 2 of those points were lost if the graph was not labeled. 2 points for finding the first positive intersection of the curves. 3 points for a setup: converting the area problem into computation of one or two definite integrals. 6 points for the subsequent computation, and 1 point of these was for the answer. The two integrals were each worth 2 points (1 for antidifferentiation and 1 for substituting the limits of integratin) and their difference was worth 1 point.

Problem 15 (12 points)
Each antidifferentiation was worth 4 points and each correct use of the initial condition earned 2 points.
2 points were deducted (one time!) if the student incorrectly declared that an antiderivative of e-t was e-t. More elaborate and incorrect antiderivatives were penalized further. 2 points were deducted if the initial values were interchanged. 1 point was deducted if somehow t's became x's (the function is x, and the independent variable here is t).