## 17Calculus - Calculus 2 - Practice Exam 3 (Semester C)

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This is the third of four exams for semester C of calculus 2.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2.5 hours

Questions

9

Total Points

70

Tools

Calculator

no

Formula Sheet(s)

1 page, 8.5x11 or A4

Other Tools

none

General Instructions:
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

Specific Instructions:
- Unless otherwise instructed, determine the convergence or divergence of these series.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test on this site unless told to use a specific test.
- You may not use the idea of growth rates as your answer when evaluating limits.

[5 points] You must successfully use each of these tests at least once to solve the rest of the problems on this exam. By successfully we mean that the test proved either convergence or divergence. It is not successful if the test was inconclusive. You must use all these tests to get any points on this question, i.e. the only two scores possible on this question are 0 and 5.

 Integral Test Ratio Test Root Test Divergence Test Direct Comparison Test Limit Comparison Test

Problem Statement

[5 points] You must successfully use each of these tests at least once to solve the rest of the problems on this exam. By successfully we mean that the test proved either convergence or divergence. It is not successful if the test was inconclusive. You must use all these tests to get any points on this question, i.e. the only two scores possible on this question are 0 and 5.

 Integral Test Ratio Test Root Test Divergence Test Direct Comparison Test Limit Comparison Test

Solution

Here are the problems we used the tests on. Yours may be different. Integral Test [question 2] Ratio Test [question 8] Root Test [question 5] Divergence Test [question 3] Direct Comparison Test [question 4] Limit Comparison Test [question 4]

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[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

Problem Statement

[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

The integral converges by the integral test.

Problem Statement

[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

Solution

In order to use the integral test, we need a continuous function. The function $$f(x)=xe^{-2x^2}$$ is a continuous function with the same values as the same points as the sum. So we can use it to integrate.
The integral test requires that we integrate $$\int_k^{\infty}{f(x)~dx}$$. We need to determine a value for k where the function is decreasing and $$f(x)$$ is positive.
$$f(x)$$ is positive when x is positive and it is decreasing. Now we need to find the largest critical value so that the function is decreasing.

 $$\displaystyle{ f(x) = \frac{x}{e^{2x^2}} }$$ $$\displaystyle{ f'(x) = \frac{e^{2x^2}(1)-xe^{2x^2}(4x)}{e^{4x^2}} }$$ $$\displaystyle{ f'(x) = \frac{1-4x^2}{e^{2x^2}} }$$

The derivative is zero when the numerator is zero and the denominator is non-zero.
$$1-4x^2=0 \to 4x^2=1 \to x^2 = 1/4 \to x = \pm1/2$$
The largest critical value is $$1/2$$. We need to make sure that the function is decreasing above this critical value. Choose $$x=1$$ as a test value.
$$\displaystyle{ f'(1) = \frac{-3}{e^2} \lt 0 }$$ so the function is decreasing and $$f(x)$$ is positive.
We choose to integrate starting at 1. You can choose any real number greater than 1/2.

 $$\displaystyle{ \int_1^{\infty}{ x e^{-2x^2} ~dx } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \int_1^{b}{ x e^{-2x^2} ~dx } } }$$

We will drop the limits of integration and integrate the indefinite integral.

 $$\displaystyle{ \int{ x e^{-2x^2} ~dx } }$$ Use integration by substitution. $$u = -2x^2 \to du = -4x~dx \to -du/4 = x~dx$$ $$\displaystyle{ \int{ e^u \frac{-du}{4} } }$$ $$\displaystyle{ \frac{-1}{4} e^u }$$ $$\displaystyle{ \frac{-1}{4} e^{-2x^2} }$$ Now apply the limits. $$\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{4} e^{-2x^2} \right]_1^b } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{4} e^{-2b^2} - \frac{-1}{4}e^{-2} \right] } }$$ $$\displaystyle{ \frac{1}{4e^2} }$$

Since the limit is finite, the integral converges, therefore the series converges by the integral test.

The integral converges by the integral test.

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[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

Problem Statement

[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

The series diverges by the divergence test.

Problem Statement

[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

Solution

 $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$ $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n+1}{n} } }$$ Let's try the divergence test. $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n+1}{n} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left[ 2 + \frac{1}{n} \right] } = 2 }$$

Since the limit is not zero, the series diverges.

The series diverges by the divergence test.

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[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

The series diverges by the direct comparison test.

Problem Statement

[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

Solution

Compare with the divergent p-series $$\sum{ 1/n }$$. If we assume the original series also diverges, we can use the direct comparison test and try to show that $$t_n \lt a_n$$.

 $$\displaystyle{ t_n = \frac{1}{n} }$$ and $$\displaystyle{ a_n = \frac{2n+1}{n^2+n} }$$ $$t_n \lt a_n$$ $$\displaystyle{ \frac{1}{n} \lt \frac{2n+1}{n^2+n} }$$ $$\displaystyle{ \frac{n^2+n}{n} \lt 2n+1 }$$ $$\displaystyle{ n+1 \lt 2n+1 }$$ $$\displaystyle{ n \lt 2n }$$ $$\displaystyle{ 1 \lt 2 }$$

The last inequality $$1 \lt 2$$ holds for all n. So the series diverges by the direct comparison test with the divergent p-series $$\sum{1/n}$$.
After we finish all the problems, we realize that we have not used the limit comparison test. The instructions do not state that we can use only one test per problem. So for this problem, we will prove it again using the limit comparison test.
We will use the same test series $$t_n=1/n$$.

 $$\displaystyle{ \lim_{n\to\infty}{ \frac{a^n}{t^n} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n+1}{n^2+n} \frac{n}{1} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n^2+n}{n^2+n} } = 2 }$$

Since the limit is greater than zero, the series diverges by the limit comparison test.

The series diverges by the direct comparison test.

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[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

The series converges by the root test.

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

Solution

Rewrite the sum $$\displaystyle{ 2 \sum_{n=2}^{\infty}{ \frac{2^{n}}{n^n} } }$$.
Try the root test.

 $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{ |a_n| } } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{ \frac{2^n}{n^n} } } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2}{n} } = 0 }$$

Since the limit is less than 1, the series converges by the root test.

The series converges by the root test.

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[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

The series converges by the Ratio Test.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

Solution

If we evaluate the limit $$\displaystyle{ \lim_{n\to\infty}{a_n} }$$, we get zero. So the divergence test won't help. Let's try the ratio test.

 $$\displaystyle{ a_n = \frac{n^2}{e^n} }$$ and $$\displaystyle{ a_{n+1} = \frac{(n+1)^2}{e^{n+1}} }$$ For the ratio test, we need to evaluate $$\displaystyle{ \lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \frac{(n+1)^2}{e^{n+1}} \frac{e^n}{n^2} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \frac{(n+1)^2}{n^2} \frac{e^n}{e^{n+1}} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \left[ 1+\frac{1}{n} \right]^2 \cdot \frac{1}{e} \right| = \frac{1}{e} < 1 }$$

Since the limit is less than one, the series converges.

The series converges by the Ratio Test.

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[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

This is a telescoping series that converges to 1/2.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

Solution

Comparing this to the series $$\sum{1/n^2}$$ using the direct comparision test or limit comparison test, we can show that this series converges. However, this looks a lot like a telescoping series, in which case we can not only determine that it converges but also what it converges to. Let's try it.

 Use partial fractions to rewrite $$a_n$$. $$\displaystyle{ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} }$$ Solving for A and B, gives us $$\displaystyle{ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} }$$

This looks promising. Let's build a table of the first few terms to see if we can see a pattern.

$$n$$

$$a_n$$

2

$$\displaystyle{ \frac{1}{2} - \frac{1}{3} }$$

3

$$\displaystyle{ \frac{1}{3} - \frac{1}{4} }$$

4

$$\displaystyle{ \frac{1}{4} - \frac{1}{5} }$$

n

$$\displaystyle{ \frac{1}{n} - \frac{1}{n+1} }$$

Okay, so notice that the nth sum is $$\displaystyle{ S_n = \frac{1}{2} - \frac{1}{n+1} }$$. If we take the limit as n goes to infinity, this will tell us what the series converges to.
$$\displaystyle{ \lim_{n\to\infty}{S_n} = \lim_{n\to\infty}{ \left[ \frac{1}{2} - \frac{1}{n+1} \right] } = \frac{1}{2} }$$

This is a telescoping series that converges to 1/2.

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[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

The series converges absolutely by the ratio test.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

Solution

Because of the factorial, the ratio test will work well.

 $$\displaystyle{ a_n = \frac{11^n}{n!} }$$ and $$\displaystyle{ a_n = \frac{11^{n+1}}{(n+1)!} }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{11^n}{n!} \frac{(n+1)!} {11^{n+1}}\right| } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{11}{n+1} \right| } = 0 < 1 }$$

Since the limit is less than one, the series converges.
In this case, we showed that $$\sum|a_n|$$ converges. So series $$\sum(-1)^{n+1}a_n$$ also converges and by the absolute convergence theorem, it converges absolutely.

The series converges absolutely by the ratio test.

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[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

This is a geometric series that converges to $$\displaystyle{ \frac{2e^{-2}}{e-2} }$$.

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

Solution

Let's rewrite this.

 $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n}} } }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \left[ \frac{2}{e} \right]^n } }$$

This looks a lot like a geometric series. We know that $$\displaystyle{ \sum_{n=0}^{\infty}{ r^n } = \frac{1}{1-r} }$$. Our sum starts at 1, so let's rewrite this as $$\displaystyle{ r^0 + \sum_{n=1}^{\infty}{ r^n } = \frac{1}{1-r} \to }$$ $$\displaystyle{ \sum_{n=1}^{\infty}{ r^n } = \frac{1}{1-r} - 1 }$$
In our problem $$r = 2/e < 1$$, so $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } = \frac{1}{e^2}\left[ \frac{1}{1-2/e} - 1 \right] }$$
Simplifying, we get $$\displaystyle{ \frac{1}{e^2}\left[ \frac{2}{e-2} \right] = \frac{2e^{-2}}{e-2} }$$

This is a geometric series that converges to $$\displaystyle{ \frac{2e^{-2}}{e-2} }$$.

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You CAN Ace Calculus

 Infinite Series

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other exams from calculus 2

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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