## 17Calculus - Calculus 2 - Practice Exam 2 (Semester B)

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This is the second and final exam for second semester single variable calculus.

### 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 hours

Questions

15

Total Points

100

Tools

Calculator

no

Formula Sheet(s)

none

Other Tools

none

Instructions:
- This exam is in four main parts, labeled sections 1-4, with different instructions for each section.
- 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).

Section 1

Evaluate each of the limits or explain why it doesn't exist. Each question in this section is worth 5 points.

$$\displaystyle{ \lim_{x\to\infty}{ \frac{\ln(x)}{x} } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to\infty}{ \frac{\ln(x)}{x} } }$$

$$\displaystyle{ \lim_{x\to\infty}{ \frac{\ln(x)}{x} } = 0 }$$

Problem Statement

$$\displaystyle{ \lim_{x\to\infty}{ \frac{\ln(x)}{x} } }$$

Solution

Direct substitution yields $$\infty/\infty$$ which is indeterminate.
So use L'Hôpital's Rule.
$$\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}=}$$ $$\displaystyle{\lim_{x \to \infty}{\frac{1/x}{1}}=}$$ $$\displaystyle{\lim_{x\to\infty}{\frac{1}{x}}=0}$$

$$\displaystyle{ \lim_{x\to\infty}{ \frac{\ln(x)}{x} } = 0 }$$

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$$\displaystyle{ \lim_{x\to 0}{ \frac{\sec(x)-1}{x^2} } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to 0}{ \frac{\sec(x)-1}{x^2} } }$$

$$\displaystyle{ \lim_{x\to 0}{ \frac{\sec(x)-1}{x^2} } = \frac{1}{2} }$$

Problem Statement

$$\displaystyle{ \lim_{x\to 0}{ \frac{\sec(x)-1}{x^2} } }$$

Solution

Direct substitution yields $$0/0$$ which is indeterminate.
So use L'Hôpital's Rule.
$$\displaystyle{\lim_{x\to 0}{\frac{\sec(x)}{x^2}}=\lim_{x\to 0}{\frac{\sec(x)\tan(x)}{2x}}=\frac{0}{0}}$$
Still indeterminate, so use L'Hôpital's Rule again.

 $$\displaystyle{ \lim_{x\to 0}{\frac{\sec(x)\tan(x)}{2x}} }$$ $$\displaystyle{\lim_{x\to 0}{\frac{\sec(x)\tan^2(x)+\sec(x)\sec^2(x)}{2}} }$$ $$\displaystyle{\frac{1}{2} }$$

$$\displaystyle{ \lim_{x\to 0}{ \frac{\sec(x)-1}{x^2} } = \frac{1}{2} }$$

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$$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{x} - \frac{1}{\sin(x)} \right] } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{x} - \frac{1}{\sin(x)} \right] } }$$

$$\displaystyle{ \lim_{x \to 0^+}{ \left[ \frac{1}{x} - \frac{1}{\sin(x)} \right]} = 0 }$$

Problem Statement

$$\displaystyle{ \lim_{x\to 0^+}{ \left[ \frac{1}{x} - \frac{1}{\sin(x)} \right] } }$$

Solution

Direct substitution yields $$\infty-\infty$$ which is indeterminate.
So use L'Hôpital's Rule. Before we can use this rule, we need the limit in fraction form.
$$\displaystyle{\lim_{x \to 0^+}{\left[ \frac{1}{x} - \frac{1}{\sin(x)}\right]} = }$$ $$\displaystyle{\lim_{x \to 0^+}{\frac{\sin(x)-x}{x\sin(x)}} = }$$ $$\displaystyle{\lim_{x \to 0^+}{\frac{\cos(x)-1}{x\cos(x)+\sin(x)}}}$$
Direct subsitution gives us $$0/0$$ which is also indeterminate. So we need to use L'Hôpital's Rule again.

 $$\displaystyle{\lim_{x \to 0^+}{\frac{\cos(x)-1}{x\cos(x)+\sin(x)}} }$$ $$\displaystyle{ \lim_{x \to 0^+}{\frac{-\sin(x)}{x(-\sin(x))+\cos(x)+\cos(x)}} }$$ $$\displaystyle{ \frac{0}{2}=0 }$$

$$\displaystyle{ \lim_{x \to 0^+}{ \left[ \frac{1}{x} - \frac{1}{\sin(x)} \right]} = 0 }$$

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Section 2

Evaluate each improper integral or show that it diverges. Each question in this section is worth 5 points.

$$\displaystyle{ \int_{0}^{\infty}{ \frac{3x}{(x^2+9)^2} dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{\infty}{ \frac{3x}{(x^2+9)^2} dx } }$$

$$\displaystyle{ \int_{0}^{\infty}{ \frac{3x}{(x^2+9)^2}dx} = \frac{1}{6} }$$

Problem Statement

$$\displaystyle{ \int_{0}^{\infty}{ \frac{3x}{(x^2+9)^2} dx } }$$

Solution

First, we need to rewrite the improper integral as a limit of a proper integral. This is important since definite integration is only defined on a continuous, closed interval. The integrand itself is continuous on the entire interval from $$0$$ to $$\infty$$, the only part of the integral that makes it improper is the upper limit.
$$\displaystyle{ \int_{0}^{\infty}{\frac{3x}{(x^2+9)^2}dx} = \lim_{b \to \infty}{\int_{0}^{b}{\frac{3x}{(x^2+9)^2}dx}}}$$
We will use integration by substitution and evaluate the indefinite integral first. Then go back, substitute the limits of integration and take the limit.
$$u = x^2+9 ~~~ \to ~~~ du = 2x~dx ~~~ \to ~~~ du/(2x) = dx$$
$$\displaystyle{\int{\frac{3x}{(x^2+9)^2}dx} =}$$ $$\displaystyle{\int{\frac{3x}{u^2}\frac{du}{2x}} =}$$ $$\displaystyle{\frac{3}{2}\int{u^{-2}du} = }$$ $$\displaystyle{ \frac{3}{2}\frac{u^{-1}}{-1} + C}$$ $$\displaystyle{= \frac{-3}{2(x^2+9)} + C}$$

 $$\displaystyle{\lim_{b \to \infty}{\int_{0}^{b}{\frac{3x}{(x^2+9)^2}dx}}}$$ $$\displaystyle{\lim_{b \to \infty}{\left[ \frac{-3}{2}\frac{1}{x^2+9} \right]_0^b}}$$ $$\displaystyle{\frac{-3}{2}\left[ \lim_{b \to \infty}{\frac{1}{b^2+9}} - \frac{1}{9} \right]}$$ $$\displaystyle{\frac{-3}{2} \left[ 0 - \frac{1}{9} \right] = \frac{1}{6}}$$

$$\displaystyle{ \int_{0}^{\infty}{ \frac{3x}{(x^2+9)^2}dx} = \frac{1}{6} }$$

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$$\displaystyle{ \int_0^2{ \frac{x}{\sqrt{4-x^2}} dx } }$$

Problem Statement

$$\displaystyle{ \int_0^2{ \frac{x}{\sqrt{4-x^2}} dx } }$$

$$\displaystyle{ \int_0^2{ \frac{x}{\sqrt{4-x^2}}dx} = 2 }$$

Problem Statement

$$\displaystyle{ \int_0^2{ \frac{x}{\sqrt{4-x^2}} dx } }$$

Solution

There is a discontinuity at $$x=2$$ since the denominator is zero. (The denominator is also zero at $$x=-2$$ but this value is not within the limits of integration. So it does not come into play in this problem.) We need to rewrite the improper integral as a limit and a proper integral.
$$\displaystyle{\lim_{b \to 2^-}{\int_0^b{\frac{x}{\sqrt{4-x^2}}dx}}}$$
As with the previous problem, we will evaluate the indefinite integral and then substitute the limits of integration and evaluate the limit later. This allows us to keep our notation correct.
$$u = 4-x^2 ~~ \to ~~ du = -2x~dx ~~ \to ~~ du/(-2x) = dx$$

 $$\displaystyle{\int{\frac{x}{\sqrt{4-x^2}}dx}}$$ $$\displaystyle{\int{\frac{x}{u^{1/2}}\frac{du}{-2x}}}$$ $$\displaystyle{\frac{-1}{2}\int{u^{-1/2}du}}$$ $$\displaystyle{\frac{-1}{2} \frac{u^{1/2}}{1/2} + C}$$ $$-(4-x^2)^{1/2} + C$$ $$\displaystyle{\lim_{b \to 2^-}{\int_0^b{\frac{x}{\sqrt{4-x^2}}dx}}}$$ $$\displaystyle{\lim_{b \to 2^-}{\left[ -(4-x^2)^{1/2}\right]_0^b}}$$ $$\displaystyle{-\lim_{b \to 2^-}{\left[ (4-b^2)^{1/2} - 2 \right]}}$$ $$\displaystyle{- \left[ (4-4)^{1/2} - 2 \right] = 2}$$

Note: There may be a tendency to write the limit as $$b \to 2$$, rather than $$b \to 2^-$$. However, we need to restrict the limit to the left of $$2$$ since the limits of integration are from $$0$$ to $$2$$. What is going on to the right of $$2$$ is not part of the integral. So it is important that the limit say $$b \to 2^-$$.

$$\displaystyle{ \int_0^2{ \frac{x}{\sqrt{4-x^2}}dx} = 2 }$$

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$$\displaystyle{ \int_0^{\infty}{ xe^{-5x} dx } }$$

Problem Statement

$$\displaystyle{ \int_0^{\infty}{ xe^{-5x} dx } }$$

$$\displaystyle{ \int_0^{\infty}{xe^{-5x}dx} = \frac{1}{25} }$$

Problem Statement

$$\displaystyle{ \int_0^{\infty}{ xe^{-5x} dx } }$$

Solution

The function is continuous on the entire interval of integration, so the only part of the limit that causes this to be an improper integral is the upper limit. So we need to rewrite this as a limit of a proper integral before we can integrate.
$$\displaystyle{\int_0^{\infty}{xe^{-5x}dx} = \lim_{b \to \infty}{\int_0^{b}{xe^{-5x}dx}}}$$
As in the previous two problems, we will evaluate the indefinite integral $$\displaystyle{ \int{xe^{-5x}dx} }$$ then substitute the limits of integration and take the limit.
Use integration by parts.

$$\to$$ $$\to$$ $$u = x$$ $$du = dx$$ $$dv = e^{-5x}dx$$ $$v = e^{-5x}/(-5)$$
 $$\displaystyle{\int{xe^{-5x}dx} }$$ $$\displaystyle{\frac{-xe^{-5x}}{5} - \int{\frac{-e^{-5x}}{5}dx} }$$ $$\displaystyle{\frac{-xe^{-5x}}{5} + \frac{1}{5}\frac{e^{-5x}}{-5} + C}$$ $$\displaystyle{\lim_{b \to \infty}{ \int_0^{b}{xe^{-5x}dx} } }$$ $$\displaystyle{\frac{-1}{5} \lim_{b \to \infty}{\left[ xe^{-5x} + \frac{1}{5}e^{-5x} \right]_0^b } }$$ $$\displaystyle{\frac{-1}{5} \lim_{b \to \infty}{\left[ be^{-5b} + \frac{1}{5}e^{-5b} - 0 - \frac{1}{5} \right]}}$$

Let's look at the first two factors. The second one is
$$\displaystyle{\lim_{b \to \infty}{ \frac{1}{5e^{5b}}} = \frac{1}{\infty} = 0}$$
The first one is
$$\displaystyle{\lim_{b \to \infty}{\frac{b}{e^{5b}}} = \frac{\infty}{\infty}}$$ which is indeterminate. So we will use L'Hôpital's Rule.
$$\displaystyle{\lim_{b \to \infty}{ \frac{b}{e^{5b}}} = }$$ $$\displaystyle{\lim_{b \to \infty}{\frac{1}{5e^{5b}}} = }$$ $$\displaystyle{\frac{1}{\infty} = 0}$$
So that leaves us with
$$\displaystyle{\frac{-1}{5} \left[ 0 + 0 - 0 - \frac{1}{5} \right] = \frac{1}{25}}$$

$$\displaystyle{ \int_0^{\infty}{xe^{-5x}dx} = \frac{1}{25} }$$

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Section 3

Determine whether these series converge or diverge. At the end of each question, state your result and which test you used to determine your answer. Each question in this section is worth 5 points.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n+100} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n+100} } }$$

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n+100} } }$$ diverges by the Divergence Test.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n+100} } }$$

Solution

Let's try the Divergence Test.
$$\displaystyle{\lim_{n \to \infty}{\left[ \frac{n}{n+100}\frac{1/n}{1/n}\right]}=}$$ $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{1}{1+100/n}\right]}=}$$ $$\displaystyle{\frac{1}{1+0}=1}$$

Since the limit is not zero, the series diverges by the Divergence Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n+100} } }$$ diverges by the Divergence Test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n\ln(n)} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n\ln(n)} } }$$

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n\ln(n)} } }$$ diverges by the integral test

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n\ln(n)} } }$$

Solution

Try the integral test. We need to evaluate $$\displaystyle{\int_{k}^{\infty}{f(x)~dx}}$$ where $$\displaystyle{f(x) = \frac{1}{x\ln(x)}}$$
We need to choose k where $$f(x)$$ is decreasing for $$x > k$$. This occurs for $$k > 1$$, so we can choose any k greater than $$1$$. We choose $$k=2$$.
So we have $$\displaystyle{\int_{2}^{\infty}{\frac{1}{x\ln(x)}dx}}$$.
If this integral diverges, so does the series; if the integral converges, so does the series. We need to write the improper integral as a limit of a proper integral.
$$\displaystyle{\int_{2}^{\infty}{\frac{1}{x\ln(x)}dx} = \lim_{b \to \infty}{\int_{2}^{b}{\frac{1}{x\ln(x)}dx}}}$$
We will evaluate the indefinite integral and then go back, evaluate the result using the limits of integration and take the limit.
Use integration by substitution and let $$u = \ln(x) ~~ \to ~~ du = dx/x ~~ \to ~~ x~du = dx$$
$$\displaystyle{\int{\frac{1}{x\ln(x)}} = }$$ $$\displaystyle{\int{\frac{1}{x}\frac{1}{u}x~du} = \int{\frac{1}{u}du} = }$$ $$\displaystyle{\ln(u)+C = \ln(\ln(x))+C}$$
$$\displaystyle{\lim_{b \to \infty}{\left[ \ln(\ln(x)) \right]_2^b} = }$$ $$\displaystyle{\lim_{b \to \infty}{[\ln(\ln(b)) - \ln(\ln(2))]} = }$$ $$\displaystyle{\infty}$$
Since the integral diverges, the series also diverges by the integral test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n\ln(n)} } }$$ diverges by the integral test

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n^2}{11n^3+4n^2+6n+3} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n^2}{11n^3+4n^2+6n+3} } }$$

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n^2}{11n^3+4n^2+6n+3} } }$$ diverges by the limit comparison test

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n^2}{11n^3+4n^2+6n+3} } }$$

Solution

Let's try the limit comparison test. We will choose the test series to be $$\sum{1/n}$$. We chose this because, if we look at the nth term of the original series for very large n, the highest powers dominate. That leaves $$\displaystyle{\frac{5n^2}{11n^3} = \frac{5}{11n} = \frac{5}{11}\cdot\frac{1}{n}}$$. Multiplication by a constant does not affect convergence or divergence, so to simplify the comparison series, we ignore $$5/11$$.
The limit comparison test says that we choose a test series and evaluate the limit $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]}}$$. If this limit turns out finite and positive (and non-zero), then the original series will converge or diverge depending on the test series. In this question, $$\displaystyle{a_n = \frac{5n^2}{11n^3+4n^2+6n+3}}$$ and $$\displaystyle{t_n = \frac{1}{n}}$$. Let's evaluate the limit.

 $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]}}$$ $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{5n^2}{11n^3+4n^2+6n+3} \frac{n}{1}\right]}}$$ $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{5n^3}{11n^3+4n^2+6n+3} \frac{1/n^3}{1/n^3} \right]}}$$ $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{5}{11+4/n+6/n^2+3/n^3} \right]}}$$ $$\displaystyle{\frac{5}{11+0+0+0} = \frac{5}{11}}$$

The limit $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]} = \frac{5}{11}}$$ and $$5/11$$ and is finite and positive. The test series is $$\sum{1/n}$$, which is a p-series with $$p=1$$ and, so, it diverges. Putting all this together tells us that the original series also diverges.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n^2}{11n^3+4n^2+6n+3} } }$$ diverges by the limit comparison test

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n+1} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n+1} } }$$

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n+1} } }$$ converges by the alternating series test

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n+1} } }$$

Solution

This is an alternating series (if you are not sure, write out the first few terms and notice how the sign alternates between positive and negative; that's the function of the $$(-1)^{n+1}$$ term). So we will use the alternating series test.
Condition 1: $$\displaystyle{\lim_{n \to \infty}{\frac{1}{n+1}} = 0}$$
So condition 1 holds.
Condition 2: We need to show that $$0 < a_{n+1} \le a_n$$
$$\displaystyle{a_n = \frac{1}{n+1} ~~~ a_{n+1} = \frac{1}{n+2}}$$
So let's test $$a_{n+1} \le a_n$$. (We can see that $$0 < a_{n+1}$$ already holds since n starts at 1 and increases.) We will set up the inequality and perform valid operations. If we get an equality that is true for all values of n, then the original inequality holds also.
$$\begin{array}{rcl} \displaystyle{\frac{1}{n+2}} & \le & \displaystyle{\frac{1}{n+1}} \\ n+1 & \le & n+2 \\ 1 & \le & 2 \end{array}$$
Since $$1 \le 2$$ is true for all $$n > 0$$ then condition 2 holds.
So the series converges by the alternating series test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}}{n+1} } }$$ converges by the alternating series test

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Section 4

Solve the following problems. Each problem in this section is worth 10 points.

Determine the convergence set of the power series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{n} } }$$

Problem Statement

Determine the convergence set of the power series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{n} } }$$

The convergence set of $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{n} } }$$ is $$-1 \leq x < 1$$.

Problem Statement

Determine the convergence set of the power series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{n} } }$$

Solution

To determine the convergence set, we need to determine the convergence interval and then evaluate the endpoints individually. For the convergence interval, we use the ratio test.
$$\displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|}}$$
For convergence, this limit must be less than one. For our question, $$\displaystyle{a_n = \frac{x^n}{n} ~~~~ a_{n+1} = \frac{x^{n+1}}{n+1}}$$
$$\begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{\left| \frac{x^{n+1}}{n+1} \frac{n}{x^n} \right|}} & < & 1 \\ \displaystyle{\lim_{n \to \infty}{\left| \frac{n}{n+1} \frac{x^{n+1}}{x^n} \right|}} & < & 1 \\ \displaystyle{\lim_{n \to \infty}{\left| \frac{n}{n+1} \frac{1/n}{1/n} x \right|}} & < & 1 \\ \displaystyle{\lim_{n \to \infty}{\left| \frac{1}{1+1/n} x \right|}} & < & 1 \\ |x| & < & 1 \end{array}$$
$$|x| < 1 ~~~ \to ~~~ -1 < x < 1$$
So the interval of convergence is $$-1 < x < 1$$. Now we need to test the endpoints.
At $$x = -1$$ the series is $$\displaystyle{\sum_{n=1}^{\infty}{\frac{(-1)^n}{n}}}$$, which converges by the alternating series test.
At $$x = 1$$, the series is $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}$$, which is a p-series with $$p=1$$ and therefore diverges.
So the convergence set of the series $$\sum{a_n}$$ is $$[-1,1)$$ (this can also be written $$-1 \leq x < 1$$).

The convergence set of $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{x^n}{n} } }$$ is $$-1 \leq x < 1$$.

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Find the power series for the function $$\displaystyle{f(x) = \frac{1}{(1-3x)^2}}$$

Problem Statement

Find the power series for the function $$\displaystyle{f(x) = \frac{1}{(1-3x)^2}}$$

Hint

$$\displaystyle{\sum_{n=0}^{\infty}{3^n x^n}}$$ is the geometric series with ratio $$r=3x$$

Problem Statement

Find the power series for the function $$\displaystyle{f(x) = \frac{1}{(1-3x)^2}}$$

$$\displaystyle{f(x) = \frac{1}{(1-3x)^2} = \sum_{n=0}^{\infty}{n(3x)^{n-1}}}$$ for $$0 < |x| < 1/3$$

Problem Statement

Find the power series for the function $$\displaystyle{f(x) = \frac{1}{(1-3x)^2}}$$

Hint

$$\displaystyle{\sum_{n=0}^{\infty}{3^n x^n}}$$ is the geometric series with ratio $$r=3x$$

Solution

A geometric series with ratio r looks like $$\displaystyle{\sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r}}$$ if $$0 < |r| < 1$$.
So in the hint, $$r=3x$$ giving us $$\displaystyle{g(x) = \sum_{n=0}^{\infty}{(3x)^n} = \frac{1}{1-3x}}$$ for $$0 < |3x| < 1 ~~ \to ~~ 0 < |x| < 1/3$$.
We can get $$f(x)$$ from $$g(x)$$ by taking the derivative of $$g(x)$$.

 $$g'(x) = \displaystyle{ (-1)(1-3x)^{-2}(-3) }$$ $$g'(x) = \displaystyle{ \frac{3}{(1-3x)^2}}$$ $$\displaystyle{\frac{g'(x)}{3}} = \displaystyle{\frac{1}{(1-3x)^2} = f(x)}$$ $$f(x) = \displaystyle{\frac{g'(x)}{3} = \frac{1}{3} \sum_{n=0}^{\infty}{3^n (nx^{x-1})}}$$ $$f(x) = \displaystyle{\sum_{n=0}^{\infty}{3^{n-1}nx^{n-1}}}$$ $$f(x) = \displaystyle{\sum_{n=0}^{\infty}{n(3x)^{n-1}}}$$

Since differentiation does not affect the convergence interval, this series converges for $$0 < |x| < 1/3$$.

$$\displaystyle{f(x) = \frac{1}{(1-3x)^2} = \sum_{n=0}^{\infty}{n(3x)^{n-1}}}$$ for $$0 < |x| < 1/3$$

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Taylors Formula with remainder applied to $$\displaystyle{f(x) = \ln(2+3x)}$$ centered about $$a=0$$ shows that $$\displaystyle{\ln(2+3x) = P_2(x) + R_2(x)}$$ where $$P_2(x)$$ is the Maclaurin polynomial of degree 2 and $$R_2(x)$$ is the remainder term. Determine $$P_2(x)$$ and $$R_2(x)$$.

Problem Statement

Taylors Formula with remainder applied to $$\displaystyle{f(x) = \ln(2+3x)}$$ centered about $$a=0$$ shows that $$\displaystyle{\ln(2+3x) = P_2(x) + R_2(x)}$$ where $$P_2(x)$$ is the Maclaurin polynomial of degree 2 and $$R_2(x)$$ is the remainder term. Determine $$P_2(x)$$ and $$R_2(x)$$.

$$\displaystyle{P_2(x) = \ln(2) + \frac{3x}{2} - \frac{9}{8}x^2}$$ and $$\displaystyle{R_2(x) = \frac{27x^2}{8(2+3x)}}$$

Problem Statement

Taylors Formula with remainder applied to $$\displaystyle{f(x) = \ln(2+3x)}$$ centered about $$a=0$$ shows that $$\displaystyle{\ln(2+3x) = P_2(x) + R_2(x)}$$ where $$P_2(x)$$ is the Maclaurin polynomial of degree 2 and $$R_2(x)$$ is the remainder term. Determine $$P_2(x)$$ and $$R_2(x)$$.

Solution

Maclaurin Polynomial
The general Taylor polynomial of degree 2 is $$\displaystyle{f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2}$$.
In our case we have a Maclaurin polynomial (a=0), our second degree polynomial is $$\displaystyle{f(0) + \frac{f'(0)}{1}(x) + \frac{f''(0)}{2}x^2}$$.

 $$f(x) = \ln(2+3x)$$ $$f(0) = \ln(2)$$ $$f'(x) = \displaystyle{\frac{3}{2+3x}}$$ $$f'(0) = 3/2$$ $$f''(x) = \displaystyle{\frac{-9}{(2+3x)^2}}$$ $$f''(0) = -9/4$$

$$\displaystyle{P_2(x) = \ln(2) + \frac{3x}{2} - \frac{9}{8}x^2}$$
Remainder Term
The general formula for the remainder term is $$\displaystyle{ R_n(x) = \frac{1}{n!} \int_{a}^{x}{(x-t)~f^{(n+1)}(t)~dt} }$$
In our case, $$n=2$$. So we need to calculate $$f^{(3)}(x)$$.
$$\displaystyle{ f^{(3)}(x) = \frac{54}{(2+3x)^3} }$$
$$\displaystyle{ R_2(x) = \frac{1}{2} \int_{0}^{x}{(x-t)f^{(3)}(t)~dt} = \frac{1}{2}\int_{0}^{x}{(x-t)\frac{54}{(2+3t)^3} dt} }$$
We can use integration by substitution with $$u = 2+3t$$.
$$u = 2+3t ~~ \to ~~ du = 3dt ~~ \to ~~ du/3 = dt$$
$$u = 2+3t ~~ \to ~~ (u-2)/3 = t ~~ \to ~~ (x-t) = (x-u/3+2/3)$$
Now let's convert the limits of integration.
$$t=0 ~~ \to ~~ u = 2$$ and $$t = x ~~ \to ~~ u = 2+3x$$

 $$R_2(x) = \displaystyle{\frac{1}{2}\int_{2}^{2+3x}{(x-u/3+2/3) \frac{54}{u^3} \frac{du}{3}}}$$ $$\displaystyle{\frac{54}{6} \int_{2}^{2+3x}{\frac{x-u/3+2/3}{u^3} ~du }}$$ $$\displaystyle{9 \int_{2}^{2+3x}{ \frac{x+2/3}{u^3} - \frac{1}{3} \frac{1}{u^2} ~du}}$$ $$\displaystyle{9 \left[ (x+2/3)\frac{u^{-2}}{-2} -\frac{1}{3} \frac{u^{-1}}{-1} \right]_{2}^{2+3x}}$$ $$\displaystyle{9 \left[ \frac{-1}{6}\frac{(2+3x)}{u^2} + \frac{1}{3}\frac{1}{u} \right]_{2}^{2+3x}}$$ $$\displaystyle{\frac{-9}{6}\left[ \frac{(2+3x)}{(2+3x)^2} - \frac{2}{(2+3x)} \right] + \frac{9}{6} \left[ \frac{(2+3x)}{4} - \frac{2}{2} \right]}$$ $$\displaystyle{\frac{3}{2} \left[ \frac{-1}{(2+3x)} + \frac{2}{2+3x} + \frac{2+3x}{4} - 1 \right]}$$ $$\displaystyle{\frac{3}{2} \left[ \frac{1}{2+3x} + \frac{2+3x}{4} - \frac{4}{4} \right]}$$ $$\displaystyle{\frac{3}{2} \left[ \frac{1}{2+3x} \frac{4}{4} + \frac{-2+3x}{4}\frac{2+3x}{2+3x} \right]}$$ $$\displaystyle{\frac{3}{2} \left[ \frac{4 - 4 + 9x^2}{4(2+3x)} \right]}$$ $$\displaystyle{\frac{27x^2}{8(2+3x)}}$$

$$\displaystyle{P_2(x) = \ln(2) + \frac{3x}{2} - \frac{9}{8}x^2}$$ and $$\displaystyle{R_2(x) = \frac{27x^2}{8(2+3x)}}$$

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Find the tangent line to the parametric curve $$x = t^3 + t, ~~~ y = t^2 -1$$; $$0 < t < 2$$ at the point $$(2,0)$$.

Problem Statement

Find the tangent line to the parametric curve $$x = t^3 + t, ~~~ y = t^2 -1$$; $$0 < t < 2$$ at the point $$(2,0)$$.

$$y = x/2 - 1$$

Problem Statement

Find the tangent line to the parametric curve $$x = t^3 + t, ~~~ y = t^2 -1$$; $$0 < t < 2$$ at the point $$(2,0)$$.

Solution

First, we need to find the value of t at the point $$(2,0)$$. We can use either the x or y expression. Using $$y = t^2-1$$ we have
$$0 = t^2 - 1 ~~ \to ~~ t^2 = 1 ~~ \to ~~ t = \pm 1$$
Since we are given the range on t as $$0 < t < 2$$, we know that $$t = 1$$.
To find the tangent line, we need the slope and a point. We are given the point $$(2,0)$$, so we need to calculate the slope.
$$\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} }$$
$$dy/dt = 2t ~~~~ dx/dt = 3t^2+1$$
$$\displaystyle{ \frac{dy}{dx} = \frac{2t}{3t^2+1} }$$
At $$t=1$$, the slope is $$m = dy/dx = 2/4 = 1/2$$
So we have the slope $$m = 1/2$$ at the point $$(2,0)$$. The equation of the tangent line can now be calculated as
$$y - 0 = (1/2)(x-2) ~~~~ y = x/2 - 1$$

$$y = x/2 - 1$$

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Calculate the area of one leaf of the 3-leaved rose $$r=5\sin(3\theta)$$.

Problem Statement

Calculate the area of one leaf of the 3-leaved rose $$r=5\sin(3\theta)$$.

Area = $$(25\pi)/4$$

Problem Statement

Calculate the area of one leaf of the 3-leaved rose $$r=5\sin(3\theta)$$.

Solution

In order to know how to set up the integral, we need to know what the graph looks like. So it is important to be able to graph polar equations on your calculator quickly. The graph we need is shown here. We can calculate the area of any of the leaves and it would be good practice for you to set up the integrals for each leaf. The general integral to calculate area is
Area = $$\displaystyle{ \frac{1}{2} \int_{\theta_1}^{\theta_2}{[ r(\theta) ]^2 ~d\theta} }$$
For this solution, we will calculate the right half of the lower leaf and multiply by $$2$$. [We do this quite often. We think it is easier to choose a leaf that is cut in half by one of the axes, calculate half and multiply by 2.] So for this question and our choice of area to calculate, our integral is

 $$\displaystyle{ \frac{1}{2} \int_{\theta_1}^{\theta_2}{[ r(\theta) ]^2 ~d\theta} }$$ $$\displaystyle{ 2 \left( \frac{1}{2} \int_{3\pi/2}^{2\pi}{ [5\sin(3\theta)]^2 ~d\theta } \right) }$$ $$\displaystyle{ 25 \int_{3\pi/2}^{2\pi}{\sin^2 (3\theta) ~ d\theta} }$$ Recall that $$\sin^2(x) = [1-\cos(2x)]/2$$. $$\displaystyle{25 \int_{3\pi/2}^{2\pi}{\frac{1}{2}(1-\cos(6\theta)) ~d\theta}}$$ $$\displaystyle{\frac{25}{2} \left[ \theta - \frac{\sin(6\theta)}{6} \right]_{3\pi/2}^{2\pi}}$$ $$\displaystyle{\frac{25}{2}[ 2\pi - 0 ] - \frac{25}{2} [ 3\pi/2 - 0 ]}$$ $$\displaystyle{25\pi - \frac{75\pi}{4}}$$ $$\displaystyle{\frac{25\pi}{4}}$$

Area = $$(25\pi)/4$$

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You CAN Ace Calculus

 l'hôpital's rule improper integrals infinite series parametrics (including calculus) polar coordinates (including calculus)

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### 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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