Partial credit in grading the first exam

Some general comments

If an indefinite integral is asked for and a "+C" is omitted, 1 point will be deducted, but only 1 time in the exam.
If a student makes a mistake early in a problem, I will try to "read with" the student and give them an appropriate amount of credit. But if a student's errors materially simplify a later part of a problem, then full credit cannot be earned for that part of the problem.

1. (14 points)
2 points for correctly writing the integrand in partial fractions
2 points for combining the pieces and equating the tops of the fractions
3 points for solving for the unknowns (1 point per unknown)
4 points for correctly integrating the resulting function
3 points for getting the correct answer by valid substitution in the functions
Merely writing that the integral, at any stage, was equal to the numerical answer without evidence does not get credit awarded.

2. (8 points)
a) 2 points for the setup: the correct integral with limits and integrand
1 point for integration
1 point for evaluation
b) 2 points for the setup: the correct integral with limits and integrand
1 point for integration
1 point for evaluation

3. (12 points)
a) 3 points. The curves should both have amplitude 1 and should both be labelled.
b) 3 points, 1 for each point.
c) 2 points for the setup of the problem (2 integrals, etc.)
2 points for antidifferentiating correctly
2 points for correct evaluation

4. (10 points)
a) 2 points for writing the average value correctly as an integral divided by the length of the appropriate interval.
2 points for the correct trig substitution allowing simple integration
2 points for correct antidifferentiation
1 point for the final answer
There certainly are other valid ways of antidifferentiation without using the trig identity (for example, using integration by parts). Student work will be read for a correct method.
b) 1 point for the answer
2 points for the verification, via L'Hospital's rule, etc.
5. (12 points)
a) 3 points for valid and relevant inequalities connecting the integrand with some simpler function.
6 points for explaining why another improper integral converges or for carefully citing a reason. "It converges" or "the function converges" is not valid.
b) 3 points for showing why the integral is less than 1/2.

6. (12 points)
a) (5 points)
2 points for converting the integral using the correct trig substitution.
1 point for the antidifferentiation.
2 points for converting the integral back into x's.
Other valid strategies could be employed. None were seen.
b) (7 points)
2 points for substituting to convert to a rational function.
2 points for correctly splitting the integrand using partial fractions.
2 points for correctly integrating the pieces.
1 point for converting back to x's.
Other valid strategies could be employed. None were seen.

7. (10 points)
2 points for a correct substitution.
4 points for integrating by parts correctly.
2 points for converting back to x's correctly, although these points do not have to be earned separately if the student decided to change the limits of integration (and this was read generously, since the limits are the same with the most common substitution).
2 points for correct substitution
There may be other valid strategies for computing this integral, although no other completely successful strategy was seen. Merely writing that the integral, at any stage, was equal to the numerical answer without evidence does not get credit awarded.

8. (12 points)
a) 6 points for a correct use of integration by parts resulting in a formula with reduced n.
b) (6 points)
2 points each for each of 2 applications of the formula in a) followed by another 2 points for the final integration of a power of x

9. (10 points)
2 points for a correct first derivative of the integrand
3 points for a correct second derivative of the integrand
3 points for identifying an overestimate of the absolute value of the second derivative on the intervbal [0,2]. Of these, 2 are reserved for a reason why the asserted overestimate is correct.
2 points for deducing an n which is a vlid answer supported by the student's computations.