About the Practice Exam 3


Problem 5:

Given that

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}{a_n}
\end{equation}

converges state which of the following converges, may converge or does not converge.

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}

We can use the limit comparison test. Let

\[ b_n=\frac{a_n}{n^2} \]

Then,

\[ \frac{b_n}{a_n} = \frac{1}{n^2} \]

Since

\begin{equation} \label{eq:poly}
\lim n\rightarrow \infty \frac{1}{n^2} =0
\end{equation}

We can conclude that

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}converges

We can also use the comparison test by noting that

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2} < \sum_{n=1}^{\infty}{a_n}
\end{equation}

to conclude that

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}\frac{a_n}{n^2}
\end{equation}converges

Problem 7

To determine the radius of convergence we can apply the ratio test to

\begin{equation} \label{eq:poly}
\sum_{n=1}^{\infty}{(n+2)}{x^n}

\end{equation}

$\frac{a_{n+1}}{a_n} = \frac{(n+3)x^{n+1}}{(n+2)x^n} =\frac{(n+3)}{(n+2)}x $

Since $\lim n\rightarrow \infty \frac{n+2}{n+1} =1$ we need that $|x|<1$ for convergence.

To get the interval of convergence we need to check the two endpoints of $-1,1$

If $x = -1$, the series will alternate and increase since $n+2$ is increasing. The same is true for $x=1$ the terms are increasing in size.

Conclusion: Interval of convergence is $(-1,1)$

Problem 10

Apply the root test to: $\sum_{n=1}^{\infty}\frac{1}{5^n}$

$\sqrt[n]{a_n}=\;\sqrt[n]{\frac{1}{5^n}}\;=\frac{1}{5}\;<1$ so converges.

Problem 12

$ln|1-x|\;=\int \frac{1}{1-x}dx$ so to get the power series for $\ln|1-x|$ we can integrate the series for $\frac{1}{1-x}$ which is

\begin{equation} \label{eq:poly}
\frac{1}{1-x}\;=\sum_{n=0}^{\infty}x^n
\end{equation}

Exam 2 Problem 6

Your classmate, Yinhe, pointed out that the part b problem actually diverges.

This is the Improper integral in question:

$latex  \int_{0}^{\infty} \frac {1}{2+sin^2 x} dx$

We can use a simple comparison test.

\[sin^2 x ≤ 1\]

\[

2 +sin^2 x  ≤ 3

\] \[

\frac {1}{2+sin^2 x} ≥ 1/3

\]

$latex  \int  \frac {1}{2+sin^2 x}  dx ≥ \int  \frac {1}{3} dx = \frac {x}{3} + C $

Since $latex \lim_{x \to \infty} \frac {x}{3} } = \infty$

$latex  \int_{0}^{\infty} \frac {1}{3} dx$ diverges

And so $latex  \int_{0}^{\infty} \frac {1}{2+sin^2 x} dx$ also diverges.

Trigonometric Substitution: Problem 4

Solve:       $latex A = \int  \frac {1}{(4+x^2)^2}  dx$

This is the form of arctan so we use the substitution

$2tan(\theta) = x

the “2” is from “4” + x^2

dx = 2  sec^2\theta d\theta

Simplify: 4+x^2 = 4(1+tan^2\theta) = 4 sec^2\theta

Substitute: A = \int \frac{2  sec^2\theta d\theta}{(4sec^2\theta)^2}

= 1/8 \int cos^2\theta d\theta

We have solved \int cos^2\theta d\theta    \qquad before.

If you work through it you should get

\int cos^2\theta d\theta = \frac{\theta}{2} + \frac{2sin\theta cos\theta}{4}

Substitute for \theta : \theta = arctan(x/2)

sin\theta = \frac{x}{\sqrt{x^2 +4}}

and    \qquad cos\theta = \frac{{2}}{\sqrt{x^2 +4}}

Prove out the final result:$

$A = \frac{ arctan( x/2)}{2} +\frac {{x}}{x^2 +4}  + C

$

Exam 1 Review

Look through these sections of the text and Webwork to get a good idea for the problems for the exam. You can review other topics as well, as they may be interwoven into some of the problems.

You should know Indefinite Integral as anti-derivative; Definite integral as Area; Fundamental Theorem of Calculus; Basic anti-derivatives; Integration techniques

Text Chapter §6.1 Problems 3-10, 24-30 (Substitution method)

§6.2 Examples 159-164 (Integration by Parts)

§6.3 Example 169 (trig)

WebWork Definite Integrals Problems 3 – 12  (basic integration)

WebWork FTC II Problems 9,10,11,13,15

Text Examples 175-178 (Trig Substitution)

Integration by Parts Problem 8

This discussion sets the analysis but it needs to be completed. Please add the finishing touches.

The formatting uses “latex”. Add this to your text:  [ latexpage]  (no space before the “l”)

Here is my latex code for the first line: (Start and end with a Dollar Sign)

Solve:      latex \int ln(x^2 + 15x + 54) dx

Solve:      $latex \int ln(x^2 + 15x + 54) dx $

Factor:

$latex x^2 + 15x + 54   =(x+6)(x+9)

\int ln(x^2 + 15x + 54) dx =  \int ln(x+6) + ln(x+9)dx

Now take one of them: \int ln(x+6)dx

Set u = ln(x+6) and dv = dx

du = dx/(x+6) and v = x\\

Result using Integration by Parts:

\int ln(x+6)dx = xln(x+6) – \int x/(x+6)dx =

xln(x+6) – \int (1 – 1/(x+6) )dx = xln(x+6) -ln|x+6| + c

$