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

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > exam list > calc2 exam B2

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

### Practice Exam Tips

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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.
$$\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}=}$$ $$\displaystyle{\lim_{x \to \infty}{\frac{1/x}{1}}=}$$ $$\displaystyle{\lim_{x\to\infty}{\frac{1}{x}}=0}$$

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

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

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.

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

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

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

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.

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

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

### 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.
$$u = x^2+9 ~~~ \to ~~~ du = 2x~dx ~~~ \to ~~~ du/(2x) = dx$$
$$\displaystyle{\int{\frac{3x}{(x^2+9)^2}dx} =}$$ $$\displaystyle{\int{\frac{3x}{u^2}\frac{du}{2x}} =}$$ $$\displaystyle{\frac{3}{2}\int{u^{-2}du} = }$$ $$\displaystyle{ \frac{3}{2}\frac{u^{-1}}{-1} + C}$$ $$\displaystyle{= \frac{-3}{2(x^2+9)} + C}$$

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

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

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

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 $$b \to 2^-$$.

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

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

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
$$\displaystyle{\frac{-1}{5} \left[ 0 + 0 - 0 - \frac{1}{5} \right] = \frac{1}{25}}$$

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

### 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.

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

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

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

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

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

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

### 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 $$-1 \leq x < 1$$.

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 $$0 < |x| < 1/3$$

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 $$\displaystyle{R_2(x) = \frac{27x^2}{8(2+3x)}}$$

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
$$y - 0 = (1/2)(x-2) ~~~~ y = x/2 - 1$$

$$y = x/2 - 1$$

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