Math 237. Calculus II
Exam 1 (non-calculator portion)
Fall, 2011
Name: Solutions
Evaluate the following limits.
This has the indeterminate form /, so L'Hpital's rule applies.
This also has the indeterminate form /, but using L'Hpital's rule at this point
Math 237. Calculus II
First Quiz, Version 1
Name: _
Give the definitions for each of the following statements.
5. The following is the definition (rephrased) for a certain kind of limit.
For every positive number, a, there is a positive number, b, such th
Math 237. Calculus II
First Quiz, Version 2
Name: _
Give the definitions for each of the following statements.
5. The following is the definition (rephrased) for a certain kind of limit.
For every positive number, a, there is a positive number, b, such th
Math 237. Calculus II
Quiz 3
Name: _
Find the following integrals.
NOTE: The "arc" notation is allowed only for the trigonometric functions. So there is no arccosh x.
Fall, 2011
Math 237. Calculus II
Quiz 4 (on 7.3)
Name: _
For each of the following nine questions, rewrite the integral using the applicable technique.
1.
Factor out cos x, then rewrite the rest.
2.
3.
4.
5.
Write sin4x = (sin2x)2, then substitute
Factor out sin x,
Math 237. Calculus II
Exam 3
Fall, 2011
Name: SOLUTIONS _
Points: All questions are worth ten points except #6 and #9, which are worth 13, and #13, which is worth 14.
Part I: Integrals. Find the following definite and indefinite integrals. If any integral
Math 237. Calculus II
Exam 4, Version 1
Fall, 2011
Name: SOLUTIONS .
1. Find the indefinite integral
Substitute u = ln x. This gives du = dx/x, and so
2. Find the convergence set for the series
.
Starting off using the Absolute Ratio Test:
So the series w
Math 237. Calculus II
Exam 4, Version 2
Fall, 2011
Name: _
1. Find the indefinite integral
Substitute u = ln x. This gives du = dx/x, and so
2. Find the convergence set for the series
Starting off using the Absolute Ratio Test:
So the series will converge
Math 237. Calculus II
Exam 1 (calculator portion)
Fall, 2011
Name: _
1. Use the bisection method to estimate the smallest positive zero of y = 2 sin x cos 2x. Show your
intermediate results by filling in the following table.
I've extended the table some s