Question:

  1. Use integration by substitution to solve the integral below:

\[\int { 4 {e}^{-7x}dx}\]

Answer:

So for this equation, we know that integrating \({e}^{x}dx\) would be much easier, so let’s do the following substitution:

\[u = - 7x\]

\[du = -7dx \rightarrow dx = \frac{du}{-7}\]

Our transformed equation looks like so:

\[\int { 4 {e}^{u}\frac{du}{-7}} = \frac{-4}{7}{e}^{u} + C = \frac{-4}{7}{e}^{-7x} + C\]

Question:

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of

\(\frac{dN}{dt}=\frac{3150}{{t}^{4}}-220\) bacteria per cubic centimeter per day, where t is the number of days since treatment began. Find a function \(N\left( t \right)\) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

Answer:

First, find the function \(N\left( t \right)\)

\[N\left( t \right) = \int{\frac{3150}{{t}^{4}}-220} = \frac{3150}{-3{x}^{3}}-220x = -\frac{1050}{{x}^{3}}-220x + C\]

So, if we set \(x = 1\) we can determine C.

\[-1050-220 + C = 6530 \rightarrow C = 7800\]

Therefore, the final function is:

\[N\left( t \right) = -\frac{1050}{{x}^{3}}-220x + 7800\]

Question:

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f\left( x \right) = 2x - 9\).

Answer:

Here, we can see that each rectangle has a width of 1 with the rectangle centers at x = [5, 6, 7, 8] so we can simply make a sum of the areas:

q3f <- function(x) {
  val <- 1 * (2 * x - 9)
  return(val)
}
q3v <- c(5, 6, 7, 8)
sum(q3f(q3v))
## [1] 16

Question:

  1. Find the area of the region bounded by the graphs of the given equations.

\[y = {x}^{2} - 2x - 2,\quad y = x + 2\]

Answer:

This can be found by subtacting the area under the top curve by the bottom curve. But first, we need to find where these curves intersect. This can be done by setting the equations equal to each other

\[{x}^{2} - 2x - 2 = x + 2\]

\[{x}^{2} - 3x - 4\]

Now, we can solve the equation:

a = 1
b = -3
c = -4

disc <- b^2 - 4 * a * c

(x1 <- (-b - sqrt(disc))/(2 * a))
## [1] -1
(x2 <- (-b + sqrt(disc))/(2 * a))
## [1] 4

So, we need to integrate over \(x =\) -1 to \(x =\) 4

\[\int _{ -1 }^{ 4 }{ {x}^{2} - 2x - 2 \quad dx} = {\left[ \frac{{x}^{3}}{3} - {x}^{2} - 2x\right]}_{-1}^{4} = -3.\overline{33}\]

Now for the next equation:

\[\int _{ -1 }^{ 4 }{x + 2 \quad dx} = {\left[ \frac{{x}^{2}}{2} + 2x \right]}_{-1}^{4} = 17.5\]

From the values at \(x = -1, x = 4\) we see that the second equation is on the top \(y = -1.5 ,y = 16\) and the first equation is on bottom \(y = 0.\overline{66}, y = -2.\overline{66}\). Therefore, we subtract the second by the first \(Area = 17.5 - (-3.33) = 20.83\)

Question:

  1. A beauty supply store expects to sell 110 flat irons during the next year. It costs $3.75 to store one flat iron for one year. There is a fixed cost of $8.25 for each order. Find the lot size and the number of orders per year that will minimize inventory costs.

Answer:

So, it is unclear how many irons go into storage, but we may be able to determine that from the number of orders we make. From here on out, \(orders = o\), \(lot\quad size = l\). We know that \(o * l = 110\). Let’s assume that the irons sell at a constant rate over the year and that we order (and recieve) the irons at the exact time needed. So each order is stored for the ratio of leftover items from an order per year (110 units). Additionally, since we are assuming a constant rate then the avrage age of an iron in storage would be the total time divided by 2. This storage time time would be \(\frac{111-o}{110*2}\), again if we assuming no shipping time. so the cost function would be:

\[Cost = 8.25 * o + 3.75 l *\frac{111 - o}{110 * 2}\]

We know that \(o*l=110\) so we can substitute for l:

\[Cost = 8.25 * o + 3.75 * \frac{110}{o} * \frac{111 - o}{110 * 2}\]

\[Cost = 8.25 * o + 1.875 * \frac{1}{o} * \left(111 - o \right)\]

\[Cost = 8.25o + \frac{208.125}{o} - 1.875\]

To get the maximum value, we can differentiate at 0:

\[\frac{d}{dCost} = 8.25 - \frac{208.125}{{o}^{2}} = 0 \rightarrow \sqrt{\frac{208.125}{8.25}} = 0\]

so 0 is 5 and o is 22

Here is a plot of our function, and derivative

Since we need to round to full numbers we have to check the whole number min and max of o. For the minimum, o is 5 and l is 22, so the actual cost is 82.88. For the maximum, o is 6 and l is 19, so the actual cost is 85.45

Question:

  1. Use integration by parts to solve the integral below.

\[\int{ln\left( 9x \right)}{x}^{6}dx\]

Answer:

So, for this equation we can see two separate equations right away as \(ln\left(9x\right)\) and \({x}^{6}\), which can be \(u\) and \(\frac{dv}{dx}\) respectively.

\[\frac{du}{dx} = \frac{\frac{d}{dx}9x}{9x} = \frac{1}{x}\]

\[{v} = \frac{{x}^{7}}{7}\]

So integration by parts is as follows:

\[\int{u\frac{dv}{dx}dx}=uv-\int{v\frac{du}{dx}dx}\]

In our case:

\[ln\left(9x\right)\frac{{x}^{7}}{7}-\int{\frac{{x}^{7}}{7}\frac{1}{x}} = \frac{ln\left(9x\right){x}^{7}}{7} - \int{\frac{{x}^{6}}{7}}\]

\[\frac{ln\left(9x\right){x}^{7}}{7} - \frac{{x}^{7}}{49} + C\]

Question:

  1. Determine whether \(f\left( x \right)\) is a probability density function on the interval \(\left[1, {e}^{6}\right]\) . If not, determine the value of the definite integral.

\[f\left( x \right) = \frac{1}{6x}\]

Answer:

So, for this to be a pdf, it would have to have a value for all x’s in the range, which is easy to verify as okay, and the integral of the range would need to be 1.

\[\int _{ 1 }^{ {e}^{6} }{\frac{1}{6x}dx}= \left[ \frac{ln\left| x \right|}{6} \right]_{1}^{{e}^{6}} = 1 - 0 = 1\] Therefore, this appears to be a probability function.