Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 17

The most general antiderivative H(theta) of the function h(theta) can be found using the following relation:
int h(theta)d theta = H(theta) + c
int (2sin theta - sec^2 theta)d theta = int (2sin theta)d theta - int (sec^2 theta)d theta
You need to use the following formulas:
intsin theta d theta = -cos theta + c => int (2sin theta)d theta = -2cos theta + c
sec^2 theta = 1/(cos^2 theta) = (tan theta)' => int sec^2 theta d theta = int (tan theta)' = tan theta + c

Gathering all the results yields:
int (2sin theta - sec^2 theta)d theta =-2cos theta - tan theta + c
Hence, evaluating the most general antiderivative of the function yields H(theta) = -2cos theta - tan theta + c .

A metal disk of radius r rotates with an angular velocity omega about a vertical axis, through a uniform field B , pointing parallel to the axis of rotation. A circuit is made by connecting one end of a resistor with resistance R to the metal axle of the disk and the other end to a sliding contact, which touches the outer edge of the disk. Find the current I in the resistor.

The speed of a point on the disk at a distance r' for 0lt= r' lt=r from the axis is v=omega r' , so the force per unit charge is f_(mag)=v xx B=omega r' B radially outward. The emf is therefore,
epsilon=int f_(mag) * dl=int_0^r omega r' B dr'=1/2 omega Br^2
The current is then
I=epsilon/R=(omega Br^2)/R
https://en.wikipedia.org/wiki/Electromotive_force

If a family falls below ____, they are considered to be living in poverty.

The federal government tracks poverty levels among the population of the United States primarily through data collected every ten years as part of the census. That data is used by the U.S. Department of Human and Human Services to draw a relatively accurate picture of the number of Americans living below the poverty line. That line is adjusted over time to account for normal inflationary factors and is also not a single line that answers the student’s question. The size of a family must be known in order to determine whether it falls below the poverty line. Attached below are links to both the Department of Human Services website and to that of the U.S. Census Bureau. Both provide data that members of the public can use to determine whether certain families are officially (for the purposes of the provision of welfare benefits) living in poverty.
In order, then, to provide an answer to the question— “if a family falls below ______ they are considered to be living in poverty”—one must know the size of the family. How many adults and children reside on the property obviously differs greatly among families. Therefore, one must consult the tables provided in the government websites linked below to answer the question. Additionally, for families living in Alaska and Hawaii, the data is adjusted to compensate for the unique socioeconomic characteristics of those two states. What the data provided in the Department of Health and Human Services shows is that the federal government considers a family of two to be living below the poverty line if its aggregate annual income is $16,460, $20,780 for a family of three, $25,100 for a family of four, and so on.
https://aspe.hhs.gov/poverty-guidelines

Intermediate Algebra, Chapter 3, 3.6, Section 3.6, Problem 36

Determine the value of $x$ for each value of $f(x)$.

a.) $f(x) = 4$

Since $f(x) = y$, we want the value of $x$ that corresponds to $y = 4$. Locate $4$ on the $y$-axis. Moving across to the graph of $f$ and down to the $x$-axis gives $x = 3$. Thus, $f(3) = 4$.


b.) $f(x) = -2$

Since $f(x) = y$, we want the value of $x$ that corresponds to $y = -2$. Locate $-2$ on the $y$-axis. Moving across to the graph of $f$ and down to the $x$-axis gives $x = 0$. Thus, $f(0) = -2$.

c.) $f(x) = 0$

Since $f(x) = y$, we want the value of $x$ that corresponds to $y = 0$. Locate on the $y$-axis. Moving across to the graph of $f$ and down to the $x$-axis gives $x = 1$. Thus, $f(1) = 0$.

Describe child labor during the industrial revolution.

I am going to assume this question is asking about the prevalence of child labor during the Industrial Revolution in England, the reasons for this and the reaction to it. In the early days of the Industrial Revolution, huge numbers of factories were built, which drew many families from country villages into the cities, where there was great demand for workers, and conditions were crowded and unhealthy. In order to afford to live near their new places of work, many adults put their children to work in factories and as chimney sweepers. While apprenticeships had always been common, never before had children as young as four been working fourteen hour days doing repetitive, low-skill tasks for a low wage. Children worked in coal mines and cotton mills, were put up chimneys, and died from the poisonous effects of the matches they sold.
Labor laws were introduced initially to constrain workers in cotton mills, but it was only stated that those between 9 and 16 could not work more than twelve hours a day. At first, these laws were not even followed. A series of labor laws were successively passed over the course of the 19th century as public reaction to child labor grew increasingly negative, but children continued to work throughout the period of heaviest industrial growth, which had a negative effect on literacy and health. Only the enforcement of the Education Act really saw child labor begin to taper off. The works of writers such as Blake and Dickens tackle this social issue.

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 60

Use a graph of $\displaystyle \int^2_0 \sin 2 \pi x \cos 5 \pi x dx$ to guess the value of the integral. Then prove that your guess is correct by using Calculus.



Based from the graph, it seems that the average value of the area is equal to 0 since the curve above the $x$-axis is the same as the curve below will cancel out since they have opposite magnitude.

Now, by using Calculus,

$
\begin{equation}
\begin{aligned}
\sin (7 \pi x) &= \sin (2 \pi x + 5 \pi x) = \sin (2 \pi x \cos (5 \pi x) + \cos (2 \pi x) \sin (5 \pi x))\\
\\
& \text{and}\\
\\
\sin (3 \pi x) &= \sin ( 5 \pi x - 2 \pi x) = \sin (5 \pi x ) \cos ( 2 \pi x) - \cos (5 \pi x ) \sin (2 \pi x)
\end{aligned}
\end{equation}
$


So,
$\sin (7 \pi x) - \sin (3 \pi x) = \sin (2 \pi x \cos (5 \pi x) + \cos (2 \pi x) \sin (5 \pi x)) - \sin (5 \pi x ) \cos ( 2 \pi x) - \cos (5 \pi x ) \sin (2 \pi x) = 2 \sin (2 \pi x) \cos (5 \pi x)$

Thus,
$\displaystyle \sin (2 \pi x) \cos (5\pi x) = \frac{\sin (7 \pi x ) - \sin (3 \pi x)}{2}$

Hence,

$
\begin{equation}
\begin{aligned}
\int^2_0 \sin (2 \pi x) \cos (5 \pi x) dx &= \int^2_0 \left[ \frac{\sin(7 \pi x) - \sin (3 \pi x)}{2} \right] dx\\
\\
&= \frac{1}{2} \left[ \frac{\cos (7\pi x)}{7 \pi} - \frac{\cos (3 \pi x)}{3 \pi} \right]^2_0
\\
\\
&= 0
\end{aligned}
\end{equation}
$

Calculus of a Single Variable, Chapter 9, 9.4, Section 9.4, Problem 16

Limit comparison test is applicable when suma_n and sumb_n are series with positive terms. If lim_(n->oo)a_n/b_n=L where L is a finite number and L>0 ,then either both series converge or both diverge.
Given series is sum_(n=1)^oo(2^n+1)/(5^n+1)
Let the comparison series be sum_(n=1)^oo2^n/5^n=sum_(n=1)^oo(2/5)^n
The comparison series sum_(n=1)^oo(2/5)^n is a geometric series with ratio r=2/5<1
A geometric series converges, if 0<|r|<1
So, the comparison series which is a geometric series converges.
Now let's use the Limit comparison test with:
a_n=(2^n+1)/(5^n+1) and b_n=2^n/5^n
a_n/b_n=((2^n+1)/(5^n+1))/(2^n/5^n)
a_n/b^n=(2^n+1)/(5^n+1)(5^n/2^n)
a_n/b^n=((2^n+1)/2^n)(5^n/(5^n+1))
a_n/b^n=(1+1/2^n)(1/(1+1/5^n))
lim_(n->oo)a_n/b_n=lim_(n->oo)(1+1/2^n)(1/(1+1/5^n))
=1>0
Since the comparison series sum_(n=1)^oo2^n/5^n converges,the series sum_(n=1)^oo(2^n+1)/(5^n+1) as well ,converges as per the limit comparison test.