Problem 1

Use integration by substitution to solve the integral below.

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

\[ \begin{align} \int 4 e^{-7x} dx &= 4 \int e^{-7x} dx \\ \\ \text{Let }u &= -7x \\ du &= -7dx \\ \frac{du}{-7} &= dx \\ \\ 4 \int e^{-7x} dx &= 4 \int \frac{1}{-7} e^u \space du \\ &= -\frac{4}{7} \int e^u \space du \\ &= -\frac{4}{7} e^u + C\\ &= -\frac{4}{7} e^{-7x} + C \end{align} \]

Problem 2

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 \(N(t)\) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

\[ \begin{align} \int -\frac{3150}{t^4}-220 \space dt &= -3150 \int t^{-4} \space dt - \int 220 \space dt \\ &= -3150 (\frac{-1}{3} t^{-3}) - (220t) + C \\ &= 1050 t^{-3} - 220t + C \\ \\ N(t) &= 1050 t^{-3} - 220t + C \\ N(1) &= 6530 \\ 1050 \cdot 1^{-3} - 220 \cdot 1 + C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \\ N(t) &= 1050 t^{-3} - 220t + 5700 \end{align} \]

Problem 3

Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x - 9\).
Fig1

Fig1

We can see from the image that \(\Delta x = 1\)

At \(x = 5, 6, 7 \text{ and } 8\) (the midpoints of the rectangles), \(y = 1, 3, 5 \text{ and } 7\) respectively. We can see this visually, but we can also calculate it with the function \(f(x) = 2x - 9\).

Therefore, the \(\text{Area} = 1(1)+3(1)+5(1)+7(1) = 16\)

Problem 4

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

\[y = x^2-2x-2\\y=x+2\]

Find intersection of the graphs of the two equations.

\[ \begin{align} x^2-2x-2 &= x+2 \\ x^2-2x-x-2-2 &= 0 \\ x^2-3x-4 &= 0 \\ (x-4)(x+1) = 0 \\ x &= 4, -1 \end{align} \]

We can also plug in any value between -1 and 4 to determine which graph is ‘higher’. So to make it easy if we plug in \(x=0\), then \((x^2-2x-2) = -2)\) and \((x+2)=2\), so…

\[ \begin{align} \int_{-1}^4 ((x+2)-(x^2-2x-2)) \space dx &= \int_{-1}^4 -x^2+3x+4 \space dx \\ &= \left.\left( -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x \right) \right|_{-1}^3 \\ &= \left( -\frac{1}{3}3^3 + \frac{3}{2}3^2 + 4 \cdot 3 \right) - \left( -\frac{1}{3}(-1)^3 + \frac{3}{2}(-1)^2 + 4(-1) \right) \\ &= \left( -9 + 13\frac{1}{2} + 12 \right) - \left( \frac{1}{3} + \frac{3}{2} - 4 \right) \\ &= 20 \end{align} \]

The region bounded by the graphs of \(y = x^2-2x-2\) and \(y=x+2\) is 20.

Just for fun, let’s graph it…

fun1 <- function(x){
    x^2-2*x-2
}
fun2 <- function(x){
    x+2
}
plot(fun1, -2, 5, col="red", ylab="y")
plot(fun2, -2, 5, col="blue", add=TRUE)
legend("top", c("x^2-2x-2", "x+2"), fill=c("red", "blue"))

Problem 5

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.

First I had to do some research to understand how to turn the costs into a formula. I found this description of ‘Holding Costs’…

Holding costs are calculated by multiplying the per-unit annual holding cost by the average level of inventory. The average inventory level is equal to the number of ordered items, divided by two.

https://www.investopedia.com/terms/c/carrying-costs.asp

Let x equal the number of orders.

The holding costs using the formula described above wold be 3.75 * ((110/x)/2).

Add the fixed cost of $8.25 for each order to get the total costs for 1 year: 8.25 x + 3.75 * ((110 / x)/2).

Once we determine the x value that minimizes cost we can just divide the 110 flat irons by the number of orders to get the lot size.

\[ \begin{align} C &= 8.25x + 3.75 \cdot \frac{\frac{110}{x}}{2} \\ &= 8.25x + \frac{412.5}{2x} \\ &= 8.25x + \frac{206.25}{x} \\ \\ C' &= 8.25 - \frac{206.25}{x^2} \\ \\ 0 &= 8.25 - \frac{206.25}{x^2} \\ 8.25x^2 &= 206.25 \\ x^2 &= \frac{206.25}{8.25} \\ x^2 &= 25 \\ x &= 5 \\ \\ \text{Lot size} &= \frac{110}{5} \\ &= 22 \end{align} \]

Double checking our math in R…

library(Deriv)
C=function(x){8.25 * x + 3.75 * ((110 / x)/2)}
Cprime <- Deriv(C)
Cprime
## function (x) 
## 8.25 - 206.25/x^2
#find the critical number
critical_num=uniroot(Cprime,c(0,110))
critical_num$root
## [1] 5
#check that it's a minimum
Cdoubleprime <- Deriv(Cprime)
Cdoubleprime(critical_num$root) > 0
## [1] TRUE

An x value of 5 orders minimizes the cost and would have a corresponding lot size of 22.

Problem 6

Use integration by parts to solve the integral below.

\[ \int \ln(9x) \cdot x^6 dx\]

\[ \begin{align} \text{Let } u &= \ln(9x) \\ du &= \frac{1}{x} \\ \\ \text{Let } dv &= x^6 dx \\ v &= \frac{x^7}{7} \\ \end{align} \]

Substitutiing this into the integration by parts formula \(\int u \space dv = uv - \int v \space du\) we get:

\[ \begin{align} \int u \space dv &= uv - \int v \space du\\ \int \ln(9x) \cdot x^6 dx &= \ln(9x) \cdot \frac{x^7}{7} - \int \frac{x^7}{7} \cdot\frac{1}{x} \\ &= \ln(9x) \cdot \frac{x^7}{7} - \int \frac{1}{7} x^6 \\ &= \ln(9x) \cdot \frac{x^7}{7} - \frac{1}{49} x^7 + C \\ &= \frac{1}{7}x^7 \ln(9x) - \frac{1}{49} x^7 + C \end{align} \]

Problem 7

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

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

\[ \begin{align} \int_1^{e^6} \frac{1}{6x} dx &= \frac{1}{6} \int_1^{e^6} \frac{1}{x} dx \\ &= \left.\frac{1}{6} \ln x \right|_1^{e^6} \\ &= \frac{1}{6} (\ln e^6 - \ln 1) \\ &= \frac{1}{6} (6- 0)\\ &= 1 \end{align} \]

Since \(\int_1^{e^6} \frac{1}{6x} dx = 1\), \(f(x)\) is a probability density function on the interval [1, e^6].