Nov 15

A big mistake on Quiz 12 has to do with the “going to zero test.” If $$\lim_{n\to\infty}a_n\ne0$$, then $$\sum_{n=1}^\infty a_n$$ does not converge. If $$\lim_{n\to\infty}a_n=0$$, then $$\sum_{n=1}^\infty a_n$$ might converge, but might not. For example, $$\sum_{n=1}^\infty 1/n$$ does not converge, even though $$\lim_{n\to\infty} 1/n=0$$.

The scores on Quiz 12 are 30,30,30,25,24,23,23,20,20,15,13,13,12,10,8,0 with μ=18.5 and σ=8.7. Individual scores are on Blackboard.

Nov 13

Topics included: error estimation for the Altrnating Series Test, absolute and conditional convergence, the Ratio and Root Tests.

Here is a solution for Extra Credit 8.

First, notice $\frac{1}{n(n+4)} = \frac{1}{n^2+4n} \lt \frac{1}{n^2}$ for $$n\ge1$$. This shows the series converges by comparing it to the $$p$$-series with $$p=2$$.

Using partial fraction decomposition, it is straightforward to show $\sum_{n=1}^\infty\frac{1}{n(n+4)} = \frac{1}{4}\sum_{n=1}^\infty\left( \frac{1}{n}-\frac{1}{n+4} \right)$ Looking at the eighth partial sum of this series reveals the telescoping pattern. \begin{multline} s_8 = \frac{1}{4}\left( \left( \frac{1}{1}-\frac{1}{5} \right) + \left( \frac{1}{2}-\frac{1}{6} \right) + \left( \frac{1}{3}-\frac{1}{4} \right) + \left( \frac{1}{4}-\frac{1}{8} \right) + \right. \\ \left. \left( \frac{1}{5}-\frac{1}{9} \right) + \left( \frac{1}{6}-\frac{1}{10} \right) + \left( \frac{1}{7}-\frac{1}{11} \right) + \left( \frac{1}{8}-\frac{1}{12} \right) \right) \\ = \frac{1}{4}\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} - \frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12} \right) \end{multline} From this it can be seen the $$n$$th partial sum for $$n\ge8$$will look like $s_n=\frac{1}{4}\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} - \frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4} \right)$ Therefore, \begin{align*} \sum_{n=1}^\infty\frac{1}{n(n+4)} &= \lim_{n\to\infty}\frac{1}{4}\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} - \frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4} \right)\\ &= \frac{1}{4}\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \right) = \frac{25}{48} \end{align*}

Nov 10

Quiz 12 may include topics through §11.4.

During the rest of the quizzes and the test, you may have a cheat sheet consisting of one $$8\frac{1}{2}\times11$$ in$$^2$$ sheet of paper with anything written on it.

Topics today included the integral test, comparison test, limit comparison test and a first peek at the alternating series test.

Nov 9

The scores on Quiz 11 are 30,28,27,25,23,22,19,17,14,12,10,10,10,9,7,5,4 with μ=16 and σ=8.5. Individual scores are on Blackboard.

Nov 8

There is apparently some terminology confusion surrounding series. A series of the form $$\sum_{n=1}^\infty a_n$$ is related to two different sequences: the sequence of its terms and the sequence of its partial sums.

The sequence of its terms is $$a_1,\ a_2,\ a_3,\dots$$.

The sequence of its partial sums is $$s_1=a_1,\ s_2=a_1+a_2,\ s_3=a_1+a_2+a_3,\dots$$.

If the sequence of terms diverges or converges to something other than $$0$$, then the sequence of partial sums cannot converge. If the sequence of terms converges to $$0$$, then the series might converge, but may not. (Remember the harmonic series $$\sum_{n=1}^\infty 1/n$$ diverges even though its sequence of terms converges to $$0$$.)

If the sequence of partial sums converges to some number $$s$$, then the series converges to $$s$$ and we write $$\sum_{n=1}^\infty a_n=s$$. In this case the sequence of terms converges to $$0$$.

After Quiz 11, the Integral Test was presented (§10.3).

Nov 6

Topics included geometric, harmonic and telescoping series as well as the “going to zero” test (§11.2).

Nov 5

Extra Credit 8. (Due November 13.) Evaluate $$\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+4)}$$.

Nov 3

Quiz 11 may include topics through §11.1.

Nov 2

The scores on Quiz 10 are 30,30,30,24,23,22,19,18,17,15,13,13,12,10,9,7,6,2 with μ=16.7 and σ=8.5. Individual scores are on Blackboard.

Oct 30

A few examples computing the areas of polar regions were done (§10.4). There was a glimpse of sequences (§11.1).

To solve Extra Credit 7, let $$\lambda(t)=\left( 2+\cos t,\sin t \right)$$ for $$0\le t\le 2\pi$$.

Revolving the circle, which is the graph of $$\lambda(t)$$, about the $$y$$-axis generates the torus we want. The surface area generated by the rotation is \begin{align*} S&=\int_0^{2\pi} 2\pi y\,ds\\ &= 2\pi\int_0^{2\pi} (2+\cos t)\sqrt{(-\sin t)^2+\cos^2 t}\,dt\\ &= 2\pi\int_0^{2\pi}(2+\cos t)\,dt\\ &= 8\pi^2 \end{align*}

Oct 27

Quiz 10 may include topics through §10.3.

There was some more discussion of polar graphing (§10.3) followed by slopes of tangents and areas inside polar curves (§10.4).

Oct 26

The integral $$\int\sec^3x\,dx$$ has come up repeatedly during the last week or so. I wrote a short note showing $\int\sec^3\,dx=\frac{1}{2}\left( \sec x \tan x + \ln|\sec x+\tan x| \right)+C$ in two different ways. This integral is then used to solve problem 28 from §8.2.

Oct 25

Polar coordinates were introduced and a variety of functions were graphed (§10.3).

The scores on Quiz 9 are 30,29,27,23,23,27,21,21,20,20,19,19,13,30,10,10 with μ=21.3 and σ=6.4. Individual scores are on Blackboard.

Oct 23

More techniques for doing calculus on parametric curves were presented (§10.2).

Here’s an animation of §10.1, #22, $$\lambda(t)=(\sin t,\cos^2 t)$$, $$-2\pi\le t\le 2\pi$$. It shows snapshots of the position of the point along the path in increments of $$\pi/10$$, with the corresponding value of $$t$$ printed in the middle. There are four complete cycles during the interval $$-2\pi\le t\le 2\pi$$. The animation just keeps repeating.

Oct 22

Extra Credit 7. (Due October 30.) A tank has the shape of a torus (doughnut) with inner radius 1 meter and outer radius 3 meters. What is its surface area?

Oct 20

Quiz 9 may include topics through §10.1.

Parametric curves were differentiated today (§10.2).

Here is a solution for Extra Credit 6.

First, notice the domain of $$f$$ is constrained by two requirements: $x-x^2\ge0 \text{ and } -1\le x\le1.$ The first requires $$0\le x\le 1$$ and the second that $$-\pi/2\le x\le\pi/2$$. Therefore, the domain of $$f$$ is $$[0,1]$$.

To determine the arc length, first compute \begin{align*} 1+\left( f'(x) \right)^2 &= 1+ \left( \frac{1-2x}{2\sqrt{x-x^2}} - \frac{1}{2\sqrt{x}}\frac{1}{\sqrt{1-x}} \right)^2\\ &= \frac{1}{1-x} \end{align*} The integral for the arc length is then \begin{align*} \int_0^1\sqrt{1+(f'(x))^2}\,dx &= \int_0^1\sqrt{\frac{1}{1-x}}\,dx\\ &= \lim_{\alpha\to1-}\int_0^\alpha \frac{dx}{\sqrt{1-x}} \qquad u=1-x\\ &= \lim_{\alpha\to1-}\int_1^{1-\alpha}\frac{-du}{\sqrt{u}}\\ &= \left.\lim_{\alpha\to1-}\left( -2\sqrt{u} \right)\right\vert_1^{1-\alpha}\\ &= \lim_{\alpha\to1-}\left( 2-\sqrt{1-\alpha} \right)\\ &= 2 \end{align*}

Oct 18

Examples of parametric functions were graphed.

The scores on Quiz 8 are 28,28,26,25,25,24,23,23,19,13,11,9,9,8,7,6,0 with μ=16.7 and σ=9.1. Individual scores are on Blackboard.

Oct 16

Topics included the surface area of revolution and an introduction to parametric functions in the plane.

Here is a solution for Extra Credit 5.

The graph of the equation looks like the following.

Solve for $$y=\pm\sqrt{x^2-x^4}$$. The graph is split into an upper and lower half.

From the obvious symmetries, it suffices to calculate the area of the region in the first quadrant and then multiply by 4. \begin{align*} 4\int_0^1 \sqrt{x^2-x^4}\,dx &= 4\int_0^1 x\sqrt{1-x^2}\,dx\qquad(u=1-x^2)\\ &= 4\int_1^0 \sqrt{u}\left( \frac{-du}{2} \right)\\ &= \left.\frac{4}{3}u^{3/2}\right\vert_0^1\\ &= \frac{4}{3} \end{align*}

Oct 14

Extra Credit 6. (Due October 20.) Find the length of the graph of $$f(x)=\sqrt{x-x^2}-\sin^{-1}\sqrt{x}$$.

Oct 13

Quiz 8 may include topics through §7.8.

The arc length formula was introduced (§8.1).

Oct 11

Here is a solution for §7.4, #36.

$\int\frac{x^4+3x^2+1}{x^5+5x^3+5x}\,dx$

After the class was over, I remembered why I had chosen this as a homework problem. There is an easy solution. Let $$u=x^5+5x^3+5x$$ so that $$du/5=(x^4+3x^2+1)\,dx$$. Then \begin{align*} \int\frac{x^4+3x^2+1}{x^5+5x^3+5x}\,dx &= \int\frac{du}{5u}\\ &= \frac{\ln |u|}{5}+C\\ &= \frac{\ln |x^5+5x^3+5x|}{5}+C \end{align*}

Solving it with partial fractions is harder, but is instructive. Here’s how you do it.

The real problem here is how to write the integrand in terms of partial fractions. First, we need to factor the denominator. $x^5+5x^3+5x=x(x^4+5x^2+5)$ To factor the degree 4 term into a product of quadratic terms, notice it is already quadratic in form, so the quadratic formula gives $x^2=\frac{-5+\sqrt{5}}{2}\text{ and }x^2=\frac{-5-\sqrt{5}}{2}.$ This shows $x^5+5x^3+5x=x\left( x^2-\frac{-5+\sqrt{5}}{2} \right)\left( x^2-\frac{-5-\sqrt{5}}{2} \right).$ To make the notation a little less cumbersome, let $p=\frac{-5+\sqrt{5}}{2} \text{ and } m=\frac{-5-\sqrt{5}}{2}$ so the factoring turns into $x^5+5x^3+5x=x\left( x^2-p \right)\left( x^2-m \right).$ Moving on to the partial fraction decomposition, the form is $\frac{x^4+3x^2+1}{x\left( x^2-p \right)\left( x^2-m \right)} = \frac{A}{x}+\frac{Bx+C}{x^2-p}+\frac{Dx+E}{x^2-m}.$ Cross multiply to get $x^4+3x^2+1 = A(x^2-p)(x^2-m)+(Bx+C)x(x^2-m)+(Cx+D)x(x^2-p).$ Multiply out the right-hand side and collect like powers of $$x$$. $x^4+3x^2+1 = (A+B+E)x^4+(C+D)x^3+(5A-Bm-Ep)x^2-(Cm+Dp)x+5A$ Equate the coefficients of powers of $$x$$. \begin{align*} A +B +E &=1\\ C +D &=0\\ 5A -Bm -Ep &=3\\ Cm +Dp &=0\\ 5A &=1 \end{align*} This system of five linear equations in five unknowns has the solution $A=\frac{1}{5},\ B=E=\frac{2}{5},\ C=D=0.$ So, \begin{align*} \int\frac{x^4+3x^2+1}{x^5+5x^3+5x}\,dx &= \int\left( \frac{1}{5x}+\frac{2x}{5(x^2-p)}+\frac{2x}{5(x^2-m)} \right)\,dx\\ &= \frac{\ln |x|}{5}+\frac{\ln |x^2-p|}{5}+\frac{\ln |x^2-m|}{5}+C\\ &= \frac{\ln |x|}{5}+\frac{\ln |(x^2-p)(x^2-m)|}{5}+C\\ &= \frac{\ln |x|}{5}+\frac{\ln |x^4+5x^2+5|}{5}+C\\ &= \frac{\ln |x^5+5x^3+5x|}{5}+C \end{align*}

The scores on Quiz 7 are 28,26,25,23,23,23,22,19,16,16,15,14,12,11,11,6,5,2 with μ=16.5 and σ=7.6. Individual scores are on Blackboard.

Oct 8

Extra Credit 5. (Due October 16.) Find the area enclosed by the graph of the curve $$x^4-x^2+y^2=0$$.

Oct 6

There will indeed be Quiz 7 on Wednesday. It may include topics through §7.5.

Here is a solution for Extra Credit 4.

Since $$\cos^2\theta=\frac{1}{2}\left( 1+\cos2\theta \right)$$, it follows that $$1+\cos x=2\cos^2\frac{x}{2}$$. Therefore, \begin{align*} \int_0^{\pi/2}\frac{\cos x\,dx}{\sqrt{1+\cos x}} &= \int_0^{\pi/2}\frac{\cos x\,dx}{\sqrt{2\cos^2\frac{x}{2}}}\\ &= \frac{1}{\sqrt{2}}\int_0^{\pi/2}\frac{\cos x\,dx}{\cos\frac{x}{2}} \end{align*} Let $$u=x/2$$ so $$x=2u$$ and $$dx=2\,du$$. \begin{align*} \phantom{\int_0^{\pi/2}\frac{\cos x\,dx}{\sqrt{1+\cos x}}} &= \sqrt{2}\,\int_0^{\pi/4}\frac{\cos2u\,du}{\cos u}\\ &= \sqrt{2}\,\int_0^{\pi/4}\frac{2\cos^2u-1}{\cos u}\,du\\ &= \sqrt{2}\,\int_0^{\pi/4}\left( 2\cos u-\sec u \right)\,du\\ &= \sqrt{2}\,\left.\left( 2\sin u-\ln|\sec u+\tan u|\vphantom{\int_{0^{\pi/4}}}\right)\right\vert_{\,\,0}^{\,\,\pi/4}\\ &= 2-\sqrt{2}\,\ln\left( \sqrt{2}+1 \right) \end{align*}

Oct 4

Improper integration was introduced (§7.8).

The scores on Quiz 6 are 30,26,26,25,24,23,17,15,14,13,12,10,9,7,7,6,1,0 with μ=14.7 and σ=9.1. Individual scores are on Blackboard.

Oct 2

There are many other clever integrations techniques (§7.5), and several were demonstrated. Sections 7.6 and 7.7 will be skipped and Wednesday we’ll move on to improper integration (§7.8).

Sep 30

Extra Credit 4. (Due October 6.) $$\displaystyle \int_0^{\pi/2}\frac{\cos x\,dx}{\sqrt{1+\cos x}}$$

Sep 29

Quiz 6 may include topics through §7.3.

Rational functions were decomposed into partial fractions (§7.4).

Here is a solution for Extra Credit 3 different from the one I did on the board today.

Substutute $$x=\sin\theta$$. \begin{align*} \int_0^1\sqrt{1+\sqrt{1-x^2}}\,dx &= \int_0^{\pi/2}\sqrt{1+\sqrt{1-\sin^2\theta}}\,\cos\theta\,d\theta\\ &= \int_0^{\pi/2}\sqrt{1+\cos\theta}\,\cos\theta\,d\theta \end{align*} Use the half-angle identity $$1+\cos\theta=2\cos^2\frac{\theta}{2}$$. \begin{align*} \phantom{\int_0^1\sqrt{1+\sqrt{1-x^2}}\,dx}&= \sqrt{2}\int_0^{\pi/2}\cos\frac{\theta}{2}\cos\theta\,d\theta \end{align*} Substitute $$\theta=2\alpha$$. \begin{align*} \phantom{\int_0^1\sqrt{1+\sqrt{1-x^2}}\,dx}&= 2\sqrt{2}\int_0^{\pi/4}\cos\alpha\,\cos2\alpha\,d\alpha\\ &= 2\sqrt{2}\int_0^{\pi/4}\cos\alpha\,(1-2\sin^2\alpha)\,d\alpha\\ &= 2\sqrt{2}\int_0^{\pi/4}\cos\alpha\,d\alpha-4\sqrt{2}\int_0^{\pi/4}\sin^2\alpha\,\cos\alpha\,d\alpha\\ &= \left.2\sqrt{2}\cos\alpha\right\vert_0^{\pi/4}-\left.4\sqrt{2}\frac{\sin^3\alpha}{3}\right|_0^{\pi/4}\\ &= \frac{4}{3} \end{align*}

Sep 27

After Quiz 5, some examples of trigonometric substitution were done.

The scores on Quiz 5 are 28,26,26,25,25,24,23,22,22,21,19,17,16,16,15,15,13,12,11,11 with μ=19.4 and σ=5.5. Individual scores are on Blackboard.

Sep 25

There was a review of the inverse trigonometric functions.

Sep 24

Extra Credit 3. (Due September 29.) $$\displaystyle \int_0^1\sqrt{1+\sqrt{1-x^2}}\,dx$$

The scores on Quiz 4 are 28,26,26,24,24,24,22,21,21,20,20,13,12,11,11,10,8,6,5,5,2,2 with μ=15.5 and σ=8.6. Individual scores are on Blackboard.

Sep 23

Here is a solution for Extra Credit 2.

Plots of all three curves are shown on the following graph.

To calculate the volume by slicing, solve the equations of the curves for $$y$$.

The calculation requires two integrals because the bottom function changes. \begin{align*} V &= \int_{-1}^0 \pi\left( \left( \frac{x+3}{2}+1 \right)^2-\left( \sqrt{-x}+1 \right)^2 \right)\,dx + \int_0^1 \pi\left( \left( \frac{x+3}{2}+1 \right)^2-\left( 2\sqrt{x}+1 \right)^2 \right)\,dx\\ &= \int_{-1}^0 \frac{\pi}{4}\left( x^2+14x-8\sqrt{-x}+21 \right)\,dx + \int_0^1 \frac{\pi}{4}\left( x^2-6x-16\sqrt{x}+21 \right)\,dx\\ &= \frac{25\pi}{6} \end{align*}
To solve it by cylindrical shells, solve the equations of the curves for $$x$$.

The calculation requires two integrals because the left function changes. \begin{align*} V &= \int_0^1 2\pi(y+1)\left( \frac{y^2}{4}+y^2 \right)\,dy + \int_1^2 2\pi(y+1)\left( \frac{y^2}{4}-(2y-3) \right)\,dy\\ &= \int_0^1 \frac{5\pi}{2}\left( y^3+y^2 \right)\,dy + \int_1^2 \frac{\pi}{2}\left( y^3-7y^2+4y+12 \right)\,dy\\ &= \frac{25\pi}{6} \end{align*}

Sep 22

Quiz 5 may include topics through §7.2.

Integrals of the form $$\int\tan^mx\sec^nx\,dx$$ were discussed (§7.2), followed by an introduction to trigonometric substitution (§7.3).

Sep 18

The question was how to do integrals of the form $$\int\sin mx\cos nx\,dx$$ and $$\int\sin^m x\cos^n x \,dx$$ (§7.2).

Sep 15

Quiz 4 may include topics through §7.1.

Integration by parts was presented.

Sep 13

Extra Credit 2. (Due September 22.) Find the volume of the solid generated by revolving about the line $$y=-1$$ the region bounded by the line $$2y=x+3$$ and outside the curves $$y^2+x=0$$ and $$y^2-4x=0$$.

The scores on Quiz 3 are 30,28,28,25,18,16,16,15,13,13,12,11,10,8,7,6,6,5,4,4,3,3,2 with μ=12.3 and σ=8.6. Individual scores are on Blackboard.

Sep 8

Quiz 3 may include topics through §6.4.

Several examples concerning work were worked (§6.4).

Don’t get into the habit of leaving the $$dx$$ off your integrals! The $$dx$$ isn’t just a placeholder; it’s there as an important part of doing substitutions.

Here is a solution for Extra Credit 1.

As seen in the figure above, it is apparent that cross sections of the wedge perpendicular to the $$x$$-axis are rectangles of height $$x\sin(\pi/6)=x/\sqrt{3}$$ and width $$2\sqrt{r^2-x^2}$$. The area of each cross section is therefore $A(x)=\displaystyle\frac{x}{\sqrt{3}} 2\sqrt{r^2-x^2}.$ Integrating this over the extent of the region gives \begin{align*} V&=\int_0^rA(x)\,dx\\ &=\int_0^r\frac{2x}{\sqrt{3}} \sqrt{r^2-x^2}\,dx\\ \end{align*} Let $$w=r^2-x^2$$ so $$-dw/2=x\,dx$$. \begin{align*} \phantom{V} &= \frac{2}{\sqrt{3}}\int_{r^2}^0\sqrt{w}\left(-\frac{dw}{2}\right)\\ &= \frac{1}{\sqrt{3}}\left.\left(\frac{2}{3}w^{3/2}\right)\right\vert_0^{r^2}\\ &= \frac{2r^3}{3\sqrt{3}}. \end{align*}

Sep 6

The Disability Resource Center is asking for a volunteer to share notes from this class with a student who has difficulty writing. Volunteers earn five service hours per credit hour of the course, plus additional service hours for completing the brief online training. Volunteers should contact Zunaira Ahmad (z0ahma01@cardmail.louisville.edu) at the Disability Resource Center, 105 Stevenson Hall.

The scores on Quiz 2 are 26,25,24,23,22,14,14,13,12,12,10,10,9,8,6,6,6,4,4,2,2 with μ=12.0 and σ=7.7. Individual scores are on Blackboard.

Sep 1

Wednesday’s Quiz 2 may include topics through §6.3.

The class period today was devoted to examples of finding volumes.

Extra Credit 1. (Due September 8.) A solid is in the shape of a wedge, cut from a round log of radius $$r$$ by two planes, one perpendicular to the axis of the log and the other intersecting the first plane at an angle of $$30^\circ$$ along a diameter of the log. What is its volume?

Aug 30

The scores on Quiz 1 are 24,23,23,21,20,17,16,15,15,14,14,13,13,12,10,10,9,8,7,6,6,2 with μ=13.5 and σ=6.1. Individual scores are on Blackboard.

After Quiz 1, several more examples of finding volumes by slicing were done (§6.2). Then the formula for finding volumes by cylindrical shells was derived (§6.3).

Aug 28

Today’s topic was volumes by slicing (§6.2). In particular, we found the formula for finding the volume of a solid of revolution.

Aug 25

Wednesday’s Quiz 1 may contain material through §6.1. Remember, this includes the review of Riemann sums.

The topic of the day was the area between curves (§6.1).

There are two categories of problems in the WebAssign assignment list, those labelled Homework and those labelled Practice. The ones with the Homework appelation are the ones that count in the WebAssign gradebook. The Practice ones don’t affect the gradebook. The Practice problems have a due date after the end of the semester as a workaround for some rigidity in the WebAssign environment.

Aug 24

If you had trouble with WebAssign yesterday, here’s why.

Dear Valued Customer,

Yesterday evening from approximately 4:45 PM ET to 6:00 PM ET and 8:30 PM ET to 11:49 PM ET students and instructors were unable to access the WebAssign platform. We sincerely apologize for this inconvenience.

We know that you and your students need reliable access to your course and learning tools and we take that responsibility seriously. Our team worked quickly to investigate and resolve the issue. As of 11:49 PM ET on August 23rd, WebAssign is now accessible and fully functional.

Our team will continue to work to prevent future interruptions from occurring to ensure that your students have the best learning experience with our platforms.

—The Cengage WebAssign Team

Aug 23

A quick review of Riemann sums (§5.2) was presented. This idea will be used several times during the next few weeks to justify some useful formulas.

If you’re working a problem on WebAssign, it might have a string of characters at the top of the window looking something like SCalcET8 5.2.002. You can translate this to a problem in the text.

SCalcET8 means it is from Stewart’s calculus text, 8th edition
5.2.002 means it is problem #2 at the end of §5.2

Aug 21

I've been getting so many emails from students telling me they won't be in class on Eclipse Day, that I've decided to move the first class period to Wednesday.

Here are some things to do in lieu of Monday’s class.

1. Read the syllabus. It is on Blackboard and can be found at the class Web site.

2. Refresh your memory of Chapter 5 from the text. In particular, we’ll be doing much with Riemann sums, the Fundamental Theorem of Calculus and antidifferentiation by substitution, so make sure you understand these topics. There are some suggested problems from the text on the class Web page you can use to test your knowledge.

3. Get squared away with WebAssign. If you are new to WebAssign, then you should watch some of the introductory videos they provide.

4. Watch the eclipse! Louisville will only get 96% obscurity of the sun, but it will still be spectacular. (Be sure to use appropriate eyewear or a pinhole camera.) There will be events surrounding the eclipse on campus at the planetarium.