# Integrating WeBWorK into Textbooks: Sample Exercises

### 1Instructive Examples1.1Arithmetic

$$7$$
Solution.
$$6 + 1 = {7}\text{.}$$

#### Checkpoint1.1.2.Declaring a Problem Seed.

$$13$$
Solution.
$$5 + 8 = {13}\text{.}$$

#### Checkpoint1.1.3.Controlling Randomness.

$$3$$
Solution.
$$1 + 2 = {3}\text{.}$$

$$x^{8}$$
Solution.
We add the exponents as follows, while including a gratuitous reference to the quadratic formula:
\begin{equation*} \begin{aligned} {x^{5}x^{3}}\amp =x^{5+3}\amp \text{Theorem 1.2.1}\\ \amp =x^{8} \end{aligned} \end{equation*}

#### Checkpoint1.1.5.Using Hints.

Hint.
Factor the number inside the radical.
$$4\sqrt{3}$$
Solution.
\begin{equation*} {\sqrt{48}}=\sqrt{4^2\cdot 3}=4\sqrt{3} \end{equation*}

#### Checkpoint1.1.6.No Randomization.

$$3$$

1.2.2.a Identify Coefficients.
$$5$$
$$-6$$
$$-8$$
Solution.
Take the coefficient of $$x^2$$ for the value of $$a\text{,}$$ the coefficient of $$x$$ for $$b\text{,}$$ and the constant for $$c\text{.}$$ In this case, they are $$a = {5}\text{,}$$ $$b = {-6}\text{,}$$ $$c = {-8}\text{.}$$
$$\left\{2,\frac{-4}{5}\right\}$$
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified $$a = {5}\text{,}$$ $$b = {-6}\text{,}$$ and $$c = {-8}\text{,}$$ so the results are:
\begin{equation*} x = {\frac{-\left(-6\right)+\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {2} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-6\right)-\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {-{\frac{4}{5}}} \end{equation*}

1.2.3.a Identify Coefficients.
1.2.3.a.i
$$6$$
Solution.
Take the coefficient of $$x^2$$ for the value of $$a\text{.}$$ In this case, $$a = {6}\text{.}$$
1.2.3.a.ii
$$-31$$
Solution.
Take the coefficient of $$x$$ for the value of $$b\text{.}$$ In this case, $$b = {-31}\text{.}$$
1.2.3.a.iii
$$-30$$
Solution.
Take the constant term for the value of $$c\text{.}$$ In this case, $$c = {-30}\text{.}$$
$$\left\{6,\frac{-5}{6}\right\}$$
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified $$a = {6}\text{,}$$ $$b = {-31}\text{,}$$ and $$c = {-30}\text{,}$$ so the results are:
\begin{equation*} x = {\frac{-\left(-31\right)+\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {6} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-31\right)-\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {-{\frac{5}{6}}} \end{equation*}

#### Checkpoint1.2.4.Copy a Problem with Tasks.

1.2.4.a Identify Coefficients.
$$2$$
$$-5$$
$$-25$$
Solution.
Take the coefficient of $$x^2$$ for the value of $$a\text{,}$$ the coefficient of $$x$$ for $$b\text{,}$$ and the constant for $$c\text{.}$$ In this case, they are $$a = {2}\text{,}$$ $$b = {-5}\text{,}$$ $$c = {-25}\text{.}$$
$$\left\{5,\frac{-5}{2}\right\}$$
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified $$a = {2}\text{,}$$ $$b = {-5}\text{,}$$ and $$c = {-25}\text{,}$$ so the results are:
\begin{equation*} x = {\frac{-\left(-5\right)+\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {5} \end{equation*}
or
\begin{equation*} x = {\frac{-\left(-5\right)-\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {-{\frac{5}{2}}} \end{equation*}

### 1.3Open Problem Library

#### Checkpoint1.3.1.Cylinder Volume.

$$360\pi\ {\rm m^{3}}$$
$$1130.97\ {\rm m^{3}}$$
Solution.
We use $$r$$ to represent the base’s radius, and $$h$$ to represent the cylinder’s height.
A cylinder’s volume formula is $$V= (\text{base area}) \cdot \text{height}\text{.}$$ A cylinder’s base is a circle, with its area formula $$A = \pi r^{2}\text{.}$$
Putting together these two formulas, we have a cylinder’s volume formula:
$$\displaystyle{ V= \pi r^{2} h }$$
Throughout these computations, all quantities have units attached, and we only show them in the final step.
1. Using the volume formula, we have:
\displaystyle{\begin{aligned} V \amp = \pi r^{2} h \\ \amp = \pi \cdot 6^{2} \cdot 10 \\ \amp = \pi \cdot 360 \\ \amp = 360 \pi \textrm{ m}^3 \end{aligned}}
Don’t forget the volume unit $$\textrm{m}^3\text{.}$$
2. To find the decimal version, we replace $$\pi$$ with its decimal value, and we have:
\displaystyle{\begin{aligned}[t] V\amp = 360 \pi \\ \amp \approx 360 \cdot 3.14\ldots \\ \amp \approx {1130.97\ {\rm m^{3}}} \end{aligned}}
Don’t forget the volume unit $$\textrm{m}^3\text{.}$$

### 1.4Antidifferentiation1.4.2WeBWorK Exercises

#### 1.4.2.1.Antiderivatives.

$$593.23432548299$$
Solution.
SOLUTION
\begin{equation*} \begin{array}{rcl} \displaystyle \int_0^{5} (4 e^x+5 \sin x)\, dx \amp =\amp \displaystyle 4 e^x-5 \cos x \Big]_0^{5} \\ \amp =\amp (4 e^{5} - 5 \cos 5) - (4 e^0 - 5 \cos0 ) \\ \amp =\amp 4 e^{5} - 5 \cos 5 + 1 \end{array} \end{equation*}

#### 1.4.2.4.

$$-\cos\!\left(x\right)+C$$

#### 1.4.2.5.

$$e^{x}+C$$

$$2x$$

### 1.6Multiple Choice

#### Checkpoint1.6.1.Drop-down/Popup.

$$\text{is not}$$
Solution.
If $$\sqrt{2}$$ were rational, then $$\sqrt{2}=\frac{p}{q}\text{,}$$ with $$p$$ and $$q$$ coprime. But then $$2q^2=p^2\text{.}$$ By the Fundamental Theorem of Arithmetic 1 , the power of $$2$$ dividing the left side is odd, while the power of $$2$$ dividing the right side is even. This is a contradiction, so $$\sqrt{2}$$ is not rational.

#### Checkpoint1.6.2.Choose one.

$$\text{The Fundamental ... of Calculus}$$
Solution.
The correct answer is The Fundamental ... of Calculus.

#### Checkpoint1.6.3.Choose a Subset of Options.

$$\text{Choice 2, Choice 4, Choice 5}$$
Solution.
The correct answer is Choice 2, Choice 4, Choice 5.

#### Checkpoint1.6.4.Choose a Subset of Options with Automated Labeling.

$$\text{B, C, D}$$
Solution.
The correct answer is B, C, D.

#### Checkpoint1.6.5.Choose a Subset of Options with Explicit Labeling.

$$\text{TACO, SUSHI, PIZZA}$$
Solution.
The correct answer is TACO, SUSHI, PIZZA.

### 1.7Tables

#### Checkpoint1.7.1.Complete this Table.

$$24$$
$$24$$
$$48$$
$$48$$
Solution.
 $$\times$$ $$6$$ $$6$$ $$4$$ $$24$$ $$24$$ $$8$$ $$48$$ $$48$$

### 1.8Graphics in Exercises

#### Checkpoint1.8.1.A static <latex-image> graph.

$$\mathop{\rm C}\nolimits\!\left(n+1,2\right)\hbox{ or }\frac{\left(n+1\right)n}{2}$$

#### Checkpoint1.8.2.A randomized <latex-image> graph.

$$48\ {\rm cm^{2}}$$

#### Checkpoint1.8.3.A <latex-image> graph affected by <latex-image-preamble>.

$$-3, 0, 3$$

#### Checkpoint1.8.5.Solve using a graph.

$$\left\{1\right\}$$
Solution.
The graph reveals that the solution set to $$f(x)=1$$ is $${\left\{1\right\}}\text{.}$$

#### Exercises

##### 1.8.1.
$$5$$
Solution.
##### 1.8.2.
$$13$$
Solution.

### 2Technical Examples2.1PGML Formatting and Verbatim Calisthenics

#### Checkpoint2.1.2.

$${\verb!<>&'";!}$$
$${\verb!#%&<>\^_`|~!}$$
$${\text{\\{\}}}$$
$$\text{ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789}$$
$${\text{!()*+,-./:=?@[]}}$$

### 2.2Subject Area Templates

#### Checkpoint2.2.1.Answer is a number or a function.

$$-8$$
$$\frac{7-8x^{7}}{x}$$
Solution.
Solution explanation goes here.

#### Checkpoint2.2.2.Answer is a function with domain issues.

$$\sqrt{x-2}$$
$$\ln\!\left(\left|\frac{x}{x-2}\right|\right)$$
Solution.
Solution explanation goes here.

#### Checkpoint2.2.3.Multiple Choice by Popup, Radio Buttons, or Checkboxes.

$$\text{Blue}$$
$$\text{Blue}$$
$$\text{Blue}$$
Solution.

#### Checkpoint2.2.4.

$$\text{Choice 3}$$
Solution.

#### Checkpoint2.2.5.Tables.

$$5$$
Solution.
The missing number is 5.

2.3.1.a
Solution.
2.3.1.b
Hint.
Solution.

### 2.5Runestone Assignment Testing

#### Exercises

##### 2.5.1.
$$2$$