You CAN Ace Calculus

Limits |

L'Hopitals Rule |

Trig Limits |

Improper Integrals |

Partial Fractions |

Infinite Series |

Polar Coordinates |

Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |

Arc Length (Vector Functions) |

Arc Length Function |

Arc Length Parameter |

Conservative Vector Fields |

Cross Product |

Curl |

Curvature |

Cylindrical Coordinates |

Lagrange Multipliers |

Line Integrals |

Partial Derivatives |

Partial Integrals |

Path Integrals |

Potential Functions |

Principal Unit Normal Vector |

Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

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Help Keep 17Calculus Free |
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This is the second exam for second semester single variable calculus.

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.

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Time | 1.5 hours |

Questions | 11 |

Total Points |

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Calculator | none |

Formula Sheet(s) | none |

Other Tools | none |

*Instructions:*

- Show all your work.

- 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).

- Give exact, simplified answers.

[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Final Answer |
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\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist |

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\)

Solution |
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Built with GeoGebra |
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\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) |

Try direct substitution first. |

\(\displaystyle{ \frac{3(-3)^2-10}{(-3)^2+2(-3)-3} = \frac{17}{0} }\) |

Since the numerator is non-zero and the denominator is zero, we have three possible answers, \(\pm\infty\) or DNE (does not exist). In order to determine which we have, we need to determine the left and right-sided limits.

It will facilitate our discussion to factor the denominator.

\(\displaystyle{ \frac{3r^2-10}{r^2+2r-3} = \frac{3r^2-10}{(r+3)(r-1)}}\)

First, let's look at the left side, i.e. \(r \lt -3\). When \(r \lt -3\) but very close to -3, the numerator is positive. The term \(r+3\) is negative and \(r-1\) is also negative.

So the signs look like \(\displaystyle{ \frac{+}{(-)(-)} \to + }\), which tells us
\(\displaystyle{ \lim_{r \to -3^-}{ \frac{3r^2-10}{r^2+2r-3} } = +\infty }\).

Now let's look at the right side of \(r=-3\). When \(r \gt -3\), the numerator remains positive and \(r-1\) term remains negative. However, since \(r \gt -3\), \(r+3\) is now positive. So the signs look like
\(\displaystyle{ \frac{+}{(+)(-)} \to - }\), which tells us
\(\displaystyle{ \lim_{r \to -3^+}{ \frac{3r^2-10}{r^2+2r-3} } = -\infty }\).

Since \(\displaystyle{ \lim_{r \to -3^-}{ \frac{3r^2-10}{r^2+2r-3} } \neq \lim_{r \to -3^+}{ \frac{3r^2-10}{r^2+2r-3} } }\), the limit \(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist.

We have included a plot of the function even though the problem statement did not ask for one.

Final Answer |
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\(\displaystyle{ \lim_{r \to -3}{ \frac{3r^2-10}{r^2+2r-3} } }\) does not exist |

close solution |

[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Final Answer |
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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } = -2 }\) |

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\)

Solution |
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Direct substitution yields, \(0/0\) which is indeterminate. So let's try L'Hopitals rule next.

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\) |

\(\displaystyle{ \lim_{x \to 0}{ \frac{(4x)\cos(2x^2)}{(-2x)e^{-x^2}} } = \frac{0}{0} }\) |

This is also indeterminate and it looks like we will not get anywhere by repeating L'Hopitals rule. So we need another strategy. We need to try to get rid of either the sine term or the exponential.

With the sine term, if we had something like \(\sin(\theta)/\theta\), the limit as \(\theta\) approaches zero is one. So let's try that.

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } }\) |

Multiply the numerator and denominator by \(2x^2\). |

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} \frac{2x^2}{2x^2} } }\) |

\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{2x^2} \frac{2x^2}{e^{-x^2}-1} } }\) |

Since \(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{2x^2} } = 1 }\) we are left with |

\(\displaystyle{ \lim_{x \to 0}{ \frac{2x^2}{e^{-x^2}-1} } }\) |

Now let's try L'Hopitals rule again. |

\(\displaystyle{ \lim_{x \to 0}{ \frac{4x}{-2x e^{-x^2}} } }\) |

Direct substitution still yields \(0/0\), However, it looks like we are getting somewhere. So use L'Hopitals rule again. |

\(\displaystyle{ \lim_{x \to 0}{ \frac{4}{-2x e^{-x^2}(-2x) + e^{-x^2}(-2)} } }\) |

Now direct substitution gives us \(4/(-2) = -2\) |

Final Answer |
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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sin(2x^2)}{e^{-x^2}-1} } = -2 }\) |

close solution |

[7 points] Evaluate \(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } }\) or show that it diverges.

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } }\) or show that it diverges.

Final Answer |
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\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \frac{\ln 3}{2} }\) |

Problem Statement |
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Solution |
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This is an improper integral with at least one issue that we need to deal with, an infinite upper limit. The lower limit looks okay since at \(x=2\), the integrand is defined. The other thing we need to check for is a problem within the limits of integration.

For this integrand, the problem points would be where the denominator is zero. So \(x^2-1=0 \to x=\pm 1\). However, both of those points are outside the limits of integration, so we do not need to consider them.

So, it looks like the only issue is the upper limit. Let's write the limit.

\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \lim_{b \to \infty}{ \int_2^b{ \frac{1}{x^2-1} dx } } }\)

Now we will evaluate the integral. Since we want to save some writing, we will drop the limits of integration for now.

\(\displaystyle{ \int{ \frac{1}{x^2-1} dx } }\) |

Since the denominator will factor, we can use partial fractions. |

\(\displaystyle{ \int{ \frac{-1/2}{x+1} + \frac{1/2}{x-1} dx } }\) |

\( (-1/2)\ln|x+1| + (1/2)\ln|x-1| \) |

Now evaluate the limit. |

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \ln|x-1| - \ln|x+1| ]_2^b } }\) |

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \ln|b-1| - \ln|b+1| - \ln|1| + \ln|3| ] } }\) |

The last two terms are constants, but we need to look closer at the first two. Direct substitution yields \(\infty - \infty\) which is indeterminate. So, to use L'Hopitals Rule, we need to a fraction.

\(\displaystyle{ \lim_{b \to \infty}{ [ \ln|b-1| - \ln|b+1| ] } }\) |

\(\displaystyle{ \lim_{b \to \infty}{ \ln \left| \frac{b-1}{b+1} \right| } }\) |

Apply L'Hopitals Rule |

\(\displaystyle{ \lim_{b \to \infty}{ \ln \left| \frac{1}{1} \right| } = 0 }\) |

So now we have \( (1/2)( 0 - 0 + \ln 3 ) = (1/2)\ln 3 \)

Final Answer |
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\(\displaystyle{ \int_2^{\infty}{ \frac{1}{x^2-1} dx } = \frac{\ln 3}{2} }\) |

close solution |

[7 points] Evaluate \(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) or show that it diverges.

Problem Statement |
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[7 points] Evaluate \(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) or show that it diverges.

Final Answer |
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\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) does not converge |

Problem Statement |
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Solution |
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\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) |

\(\displaystyle{ \lim_{b \to \infty}{ \int_{\pi}^{b}{ \sin(2x) dx } } }\) |

\(\displaystyle{ \lim_{b \to \infty}{ [\cos(2x)/2]_{\pi}^b } }\) |

\(\displaystyle{ (1/2)\lim_{b \to \infty}{ [ \cos(2b) - \cos(2\pi) ] } }\) |

\( (1/2)\displaystyle{ \lim_{b \to \infty}{ \cos(2b) } } - (1/2) \) |

The limit in the last equation does not converge.

Final Answer |
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\(\displaystyle{ \int_{\pi}^{\infty}{ \sin(2x) dx } }\) does not converge |

close solution |

[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Problem Statement |
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[14 points] Find the convergence set for the power series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(x+1)^k } }\)

Hint |
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The convergence set is what 17calculus calls the *interval of convergence*, which includes information about the endpoints.

Problem Statement |
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Final Answer |
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The convergence set is \((-4,2)\). |

Problem Statement |
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Hint |
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The convergence set is what 17calculus calls the *interval of convergence*, which includes information about the endpoints.

Solution |
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First we will use the Ratio Test to determine the *radius* of convergence and then check the endpoints to get the *interval* of convergence.

\(\displaystyle{ \lim_{k \to \infty}{ \left| \frac{a_{k+1}}{a_k} \right| } < 1 }\) |

\(\displaystyle{ a_k = \frac{k^2}{3^k}(x+1)^k }\) |

\(\displaystyle{ a_k = \frac{(k+1)^2}{3^{k+1}}(x+1)^{k+1} }\) |

\(\displaystyle{ \lim_{k \to \infty}{ \left| \left[ \frac{(k+1)^2(x+1)^{k+1} }{3^{k+1}} \right] \left[ \frac{3^k}{k^2(x+1)^k} \right] \right| } < 1 }\) |

\(\displaystyle{ \lim_{k \to \infty}{ \left| \frac{(k+1)^2}{k^2} \frac{3^k}{3^{k+1}} \frac{(x+1)^{k+1}}{(x+1)^k} \right| } < 1 }\) |

\(\displaystyle{ \lim_{k \to \infty}{ \left| \left[ \frac{k+1}{k} \right]^2 \frac{1}{3} (x+1) \right| } < 1 }\) |

\(\displaystyle{ \left| \frac{x+1}{3} \right| < 1 }\) |

\(\displaystyle{ \left| x+1 \right| < 3 }\) |

\( -3 \lt x+1 \lt 3 \) |

radius of convergence \( -4 \lt x \lt 2 \) |

Now we need to check the endpoints. |

For \(x=-4\), we have \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(-3)^k } }\) |

\(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k k^2 } }\) |

Since \(\displaystyle{ \lim_{k \to \infty}{(-1)^k k^2} \neq 0 }\) the series diverges by the divergence test. |

For \(x=2\), we have \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{k^2}{3^k}(3)^k } }\) |

\(\displaystyle{ \sum_{k=1}^{\infty}{ k^2 } }\) |

Since \(\displaystyle{ \lim_{k \to \infty}{k^2} \neq 0 }\) the series diverges by the divergence test. |

Final Answer |
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The convergence set is \((-4,2)\). |

close solution |

[10 points] Given the power series \(\displaystyle{ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots }\) for \(-1 \lt t \lt 1 \),

find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Problem Statement |
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[10 points] Given the power series \(\displaystyle{ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots }\) for \(-1 \lt t \lt 1 \),

find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Final Answer |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\) |

Problem Statement |
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find a power series for \(\displaystyle{ \frac{x^2}{x^2+3} }\).

Solution |
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We need to get our power series \(\displaystyle{ \frac{x^2}{x^2+3} }\) into the form \(\displaystyle{ \frac{1}{1-t} }\).

\(\displaystyle{ \frac{x^2}{x^2+3} }\) |

\(\displaystyle{ \frac{x^2}{3+x^2} }\) |

\(\displaystyle{ \frac{-x^2}{-3-x^2} \frac{-1/3}{-1/3} }\) |

\(\displaystyle{ \frac{x^2/3}{1+x^2/3} }\) |

\(\displaystyle{ \frac{x^2/3}{1-(-x^2/3)} }\) |

Comparing the last equation with \(\displaystyle{ \frac{1}{1-t} }\), we can see that \(t=-x^2/3\). Now use that in the series equation to get

\(\displaystyle{ (x^2/3)\frac{1}{1-(-x^2/3)} }\) |

\(\displaystyle{ \frac{x^2}{3} \left[ 1 + \frac{-x^2}{3} + \left( \frac{-x^2}{3} \right)^2 + \left( \frac{-x^2}{3} \right)^3 + \ldots \right] }\) |

\(\displaystyle{ \frac{x^2}{3} - \frac{x^4}{3^2} + \frac{x^6}{3^3} - \frac{x^8}{3^4} + \ldots }\) |

Although this answer is probably sufficient, we can write it in a more compact form.

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\)

Final Answer |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n-1} x^{2n}}{3^n} } }\) |

close solution |

[4 points] Write a geometric series that converges to 5.

Problem Statement |
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[4 points] Write a geometric series that converges to 5.

Final Answer |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\) |

Problem Statement |
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[4 points] Write a geometric series that converges to 5.

Solution |
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A geometric series that converges is
\(\displaystyle{ \sum_{n=0}^{\infty}{ r^n } = \frac{1}{1-r} }\) where \( 0 < |r| < 1 \).

If we want this equal to 5, we need solve \(\displaystyle{ \frac{1}{1-r} = 5 }\) for *r*.

\(\begin{array}{rcl}
\displaystyle{ \frac{1}{1-r} } & = & 5 \\
\displaystyle{ \frac{1}{5} } & = & 1-r \\
r & = & 1 - \displaystyle{ \frac{1}{5} } \\
r & = & \displaystyle{ \frac{4}{5} }
\end{array}\)

\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\) is the series and, since \( 0 < |r=4/5| < 1 \) holds, the series converges.

Final Answer |
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\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{4}{5} \right]^n } }\) |

close solution |

[7 points] Determine whether the series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k \frac{2k}{k^2+1} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Problem Statement |
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[7 points] Determine whether the series \(\displaystyle{ \sum_{k=1}^{\infty}{ (-1)^k \frac{2k}{k^2+1} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Final Answer |
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The series converges by the Alternating Series Test. |

Problem Statement |
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Solution |
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Since we obviously have an alternating series, we will start with the alternating series test.

*Condition 1* - \(\displaystyle{ \lim_{k \to \infty}{ a_k } = 0 }\) must hold

For this problem, \(\displaystyle{ a_k = \frac{2k}{k^2+1} }\). The limit
\(\displaystyle{ \lim_{k \to \infty}{ \frac{2k}{k^2+1} } = 0 }\) holds. So condition 1 holds.

*Condition 2* - \( 0 \lt a_{k+1} \leq a_k \) must hold

\(\displaystyle{ a_{k+1} = \frac{2(k+1)}{(k+1)^2+1} }\)

Since \(k \gt 0 \), then the first part of the inequality holds, i.e. \( 0 \lt a_{k+1} \). Now we to see if the second part holds.

\(a_{k+1} \leq a_k \) |

\(\displaystyle{ \frac{2(k+1)}{(k+1)^2+1} \leq \frac{2k}{k^2+1} }\) |

\( (k+1)(k^2+1) \leq k[ (k+1)^2 + 1 ] \) |

\( k^3 + k + k^2 +1 \leq k^3 + 2k^2 + 2k \) |

\( 1 \leq k^2+k \) |

The last inequality holds for \(k \geq 1 \). If there any question in your mind, you can take the derivative \(2k+1\) and notice that it is always positive. So the values \(k^2+k\) are increasing. And since at \(k=1\) we have \(1^2+1 =2\), the values of \(k^2+k\) are always greater than 1.

Therefore, condition 2 holds as well. Since both conditions hold, the series converges by the alternating series test.

Final Answer |
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The series converges by the Alternating Series Test. |

close solution |

[7 points] Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^2} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Problem Statement |
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[7 points] Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^2} } }\) converges or diverges. As part of your final answer, indicate which test(s) you use.

Final Answer |
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The series converges by the direct comparison test. |

Problem Statement |
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Solution |
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This problem is solved in practice problem 162 on the direct comparision test page.

Final Answer |
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The series converges by the direct comparison test. |

close solution |

[16 points] **(a)** Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).

**(b)** Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Problem Statement |
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[16 points] **(a)** Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).

**(b)** Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Hint |
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You may use the formula \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\) for some number *c* between *a* and *x*.

Problem Statement |
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**(a)** Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).

**(b)** Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Final Answer |
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(a) \( P_2(x) = (x-1) - (1/2)(x-1)^2 \) |

Problem Statement |
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**(a)** Find \(P_2(x)\), the Taylor polynomial of order 2 based at 1 for \(f(x)=\ln x\).

**(b)** Bound the error \(\displaystyle{ |R_2(x)| = | \ln x - P_2(x) | ~~\text{if}~~ \frac{1}{2} \leq x \leq \frac{3}{2} }\)

Hint |
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You may use the formula \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\) for some number *c* between *a* and *x*.

Solution |
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**(a)** The second order Taylor polynomial is given by

\(\displaystyle{ P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 }\)

Let's build a table with the values that we need and then plug them into this equation for \(a=1\).

\( f(x) = \ln x \) |
\( f(1) = \ln 1 = 0 \) |

\( f'(x) = 1/x \) |
\(f'(1) = 1/1 = 1 \) |

\( f''(x) = (-1)x^{-2} \) |
\( f''(1) = -1 \) |

So now our Taylor polynomial is \( P_2(x) = 0 + 1(x-1) + (-1/2)(x-1)^2 \)

Simplifying this a bit gives us \( P_2(x) = (x-1) - (1/2)(x-1)^2 \)

Final Answer - Part (a) |
\( P_2(x) = (x-1) - (1/2)(x-1)^2 \) |

**(b)** For this part of the solution, we use the hint \(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} }\). The bound is given by taking the absolute value of this expression, i.e. \(\displaystyle{ |R_n(x)| = \left| \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \right| }\). In our case \(n=2\), so we have \(\displaystyle{ |R_2(x)| = \left| \frac{f^{(3)}(c)}{3!} (x-1)^{3} \right| }\).

The derivative we need is \(f^{(3)}(x) = -2x^{-3} \), so we have

\(\displaystyle{ |R_2(x)| = \left| \frac{-2(c^{-3})}{6} (x-1)^{3} \right| }\)

Simplifying a bit gives us our final answer.

Note: It is not clear from the problem statement if you are asked to stop here or to determine the upper bound on the error by finding \(max{|1/(3c^3)|}\). So it would be good to ask your instructor. If they want you to find the maximum on the given interval, it occurs when \(c=1/2\), so your expression would be \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\).

Final Answer - Part (b) |
\(\displaystyle{ |R_2(x)| = \left| \frac{-1}{3c^3} (x-1)^{3} \right| }\) or \(\displaystyle{ |R_2(x)| = \frac{8}{3} \left|(x-1)^{3} \right| }\) |

Final Answer |
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(a) \( P_2(x) = (x-1) - (1/2)(x-1)^2 \) |

close solution |

[14 points] **(a)** Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.

**(b)** Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Problem Statement |
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[14 points] **(a)** Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.

**(b)** Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Final Answer |
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(a) circle of radius 1 centered at \((0,1)\) |

Problem Statement |
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**(a)** Name the curve with polar equation \(r=2\sin\theta\) and sketch the graph.

**(b)** Set up and evaluate an integral in polar coordinates for the area inside the graph of part (a).

Solution |
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Built with GeoGebra |
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**(a)** If you don't automatically know what this graph is, you can convert to rectangular coordinates and it is easy to see what the plot should look like from the resulting equation.

\( r = 2\sin \theta \) |

Multiply both sides by |

\( r^2 = 2r\sin\theta \) |

On the left we use \( r^2 = x^2+y^2 \). On the right we use \(y=r\sin\theta\). |

\(x^2+y^2 = 2y \) |

\( x^2 + y^2 - 2y = 0 \) |

Complete the square on the y-term. |

\( x^2 + y^2 - 2y + 1 - 1 = 0 \) |

\( x^2 + (y-1)^2 = 1 \) |

From the last equation, it is easy to see that we have a circle centered at \((0,1)\) with radius 1.

**(b)** The area of a circle is \(A = \pi r^2\). In this case, \(A = \pi\). This helps us check our answer.

\(\displaystyle{ A = \frac{1}{2}\int_{\theta_1}^{\theta_2}{ r^2 ~ d\theta } }\) |

For this problem, we will find the area of the right half of the circle and multiply by 2. |

\(\displaystyle{ A = (2)\frac{1}{2}\int_{0}^{\pi/2}{ (2\sin\theta)^2 ~ d\theta } }\) |

\(\displaystyle{ A = \int_0^{\pi/2}{ 4\sin^2\theta ~ d\theta } }\) |

\(\displaystyle{ A = 4 \int_0^{\pi/2}{ \frac{1-\cos(2\theta)}{2} ~ d\theta } }\) |

\(\displaystyle{ A = 2 \int_0^{\pi/2}{ 1-\cos(2\theta) ~ d\theta } }\) |

\(\displaystyle{ A = 2\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi/2} }\) |

\(\displaystyle{ A = 2 \left[ \frac{\pi}{2} - \frac{\sin \pi}{2} \right] = \pi }\) |

Final Answer |
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(a) circle of radius 1 centered at \((0,1)\) |

close solution |