Task 1:

Use integration by substitution to solve the integral below.

\(\int4e^{-7x}dx\)

Solution 1:

Let \(u=-7x\). Then \(du=-7dx\)

Now, \(\int4e^{-7x}dx=4\int e^{-7x}dx=-\frac{4}{7}\int e^{-7x}(-7)dx=-\frac{4}{7}\int e^udu=-\frac{4}{7}e^u+C=-\frac{4}{7}e^{-7x}+C\)

Final Answer: \(\int4e^{-7x}dx=-\frac{4}{7}e^{-7x}+C\)

Task 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.

Solution 2:

Since \(\frac{dN}{dt}=-\frac{3150}{t^4}-220\)…

\(N(t)=\int-\frac{3150}{t^4}-220\,dt\\N(t)=-10\int315t^{-4}+22\,dt\\N(t)=-10(\frac{315t^{-3}}{-3}+22t+C)\\N(t)=-10(-\frac{105}{t^3}+22t+C)\\N(t)=\frac{1050}{t^3}-220t-10C\)

However, we know that \(N(1)=6530\), so we can solve for C:

\(6530=\frac{1050}{(1)^3}-220(1)-10C\\6530=1050-220-10C\\6530=830-10C\\5700=-10C\\-570=C\)

By substituting for C, we get:

\(N(t)=\frac{1050}{t^3}-220t-10C\\N(t)=\frac{1050}{t^3}-220t-10(-570)\\N(t)=\frac{1050}{t^3}-220t+5700\)

Final answer: \(N(t)=\frac{1050}{t^3}-220t+5700\)

Task 3:

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

Solution 3:

The line has an \(x\)-intercept of 4.5, so each rectangle has a width of 1. The heights of the rectangles from left to right are 1, 3, 5, and 7, and since each rectangle has a width of 1 these are also the areas of the rectangles. The total area is therefore \(1+3+5+7=16\).

We could calculate this area using integration as follows:

\(A=\int_{4.5}^{8.5}2x-9\,dx\\A=_{4.5}^{8.5}[x^2-9x]\\A=[(8.5^2-9(8.5))-(4.5^2-9(4.5))]\\A=[(72.25-76.5)-(20.25-40.5)]\\A=[-4.25-(-20.25)]\\A=16\)

Checking the solution with R:

f_x_3 <- function(x) {2*x - 9}

as.numeric(integrate(f_x_3, lower = 4.5, upper = 8.5)[1])
## [1] 16

Task 4:

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

\(y=x^2-2x-2\), \(y=x+2\)

Solution 4:

To find the points where the graphs intersect, we solve the equation \(x^2-2x-2=x+2\), which results in the quadratic equation \(x^2-3x-4=0\), which has solutions \(x=-1\) and \(x=4\). Substituting these \(x\) values back into the second equation results in \(y=-1+2=1\) and \(y=4+2=6\), so these graphs intersect at \((-1,1)\) and \((4,6)\). Between those two intersections, the graph of \(y=x+2\) will be above the graph of \(y=x^2-2x-2\), so to find the area between the graphs we need to find the area under \(y=x+2\) on that interval and subtract the area under \(y=x^2-2x-2\).

Our solution then is…

\(A=\int_{-1}^4 x+2\,dx-\int_{-1}^4 x^2-2x-2\,dx\\A={\displaystyle _{-1}^{4}}[\frac{x^2}{2}+2x]-{\displaystyle _{-1}^{4}}[\frac{x^3}{3}-\frac{2x^2}{2}-2x]\\A=[(\frac{(4)^2}{2}+2(4))-(\frac{(-1)^2}{2}+2(-1))]-[(\frac{(4)^3}{3}-\frac{2(4)^2}{2}-2(4))-(\frac{(-1)^3}{3}-\frac{2(-1)^2}{2}-2(-1))]\\A=[(16)-(\frac{-3}{2})]-[(-\frac{8}{3})-(\frac{2}{3})]\\A=\frac{35}{2}-(-\frac{10}{3})\\A=\frac{105}{6}+\frac{20}{6}\\A=\frac{125}{6}\)

Final answer: The area of the region is \(\frac{125}{6}\) square units.

Checking the solution with R:

f_x_41 <- function(x) {x^2 - 2*x - 2}
f_x_42 <- function(x) {x + 2}

as.numeric(integrate(f_x_42, lower = -1, upper = 4)[1]) - as.numeric(integrate(f_x_41, lower = -1, upper = 4)[1])
## [1] 20.83333

Task 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.

Solution 5:

Assume that all lots are the same size.

Let L = Lot Size

Let N = Number of Orders

Let C = Inventory Costs

Clearly, \(N \cdot L=110\), so \(N=\frac{110}{L}\) and \(L=\frac{110}{L}\)

The fixed per-order costs will total \(8.25 \cdot N=8.25N\) or \(8.25 \cdot \frac{110}{L}=907.5L^{-1}\)

If we further assume that the orders are evenly spaced throughout the year, then the storage costs for \(N\) orders of lot size \(L\) will be \(L \cdot \frac{3.75}{N} \cdot N = 3.75L\) (Since \(N \neq 0\) as there cannot be 0 orders).

Therefore, \(C=3.75L+907.5L^{-1}\).

We can find the lot size that minimizes costs by finding \(C'\) and setting it equal to 0, then solving for L, as follows:

\(C'=3.75-907.5L^{-2}\\0=3.75-907.5L^{-2}\\907.5L^{-2}=3.75\\907.5=3.75L^2\\242=L^2\\15.55635 \approx L\)

However, lot sizes must be a whole number of flat irons, so we round our answer to the nearest whole number and conclude that the lot size that minimizes costs is 16, resulting in 6.875 orders (six full size orders and one slightly smaller order).

We can confirm this finding methodically using R:

lot_size <- c(1:110)
number_orders <- 110 / lot_size
inv_costs <- (3.75 * lot_size) + (907.5 / lot_size)
costs_table <- data.frame(lot_size, number_orders, inv_costs)

library(tidyverse)
## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr     1.1.2     ✔ readr     2.1.4
## ✔ forcats   1.0.0     ✔ stringr   1.5.0
## ✔ ggplot2   3.4.4     ✔ tibble    3.2.1
## ✔ lubridate 1.9.2     ✔ tidyr     1.3.0
## ✔ purrr     1.0.1     
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
minimum_cost <- costs_table %>%
  filter(inv_costs == min(inv_costs))

library(knitr)
kable(minimum_cost, caption = "Minimum Inventory Cost", col.names = c("Lot Size", "Number of Orders", "Inventory Costs"), align = "ccc")
Minimum Inventory Cost
Lot Size Number of Orders Inventory Costs
16 6.875 116.7188

Task 6:

Use integration by parts to solve the integral below.

\(\int \text{ln}(9x) \cdot x^6dx\)

Solution 6:

Let \(u=\text{ln}(9x)\) and \(v=x^6\). Then \(u'=\text{ln}(9x)'=\frac{1}{x}\)

Now, \(\int v\,dx=\int x^6\,dx=\frac{x^7}{7}\)

Therefore, \(\int \text{ln}(9x) \cdot x^6dx=\text{ln}(9x)\frac{x^7}{7}-\int \frac{1}{x}\frac{x^7}{7}\,dx=\text{ln}(9x)\frac{x^7}{7}-\frac{1}{7}\int x^6\,dx=\text{ln}(9x)\frac{x^7}{7}-\frac{1}{7}\frac{x^7}{7}+C=\frac{x^7 \cdot \text{ln}(9x)}{7}-\frac{x^7}{49}+C\)

Final Answer: \(\frac{x^7 \cdot \text{ln}(9x)}{7}-\frac{x^7}{49}+C\)

Task 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.

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

Solution 7:

For \(f(x)=\frac{1}{6x}\) to be a pdf on the interval \([1,e^6]\) it must meet two criteria:

  1. \(f(x)>0\) on the entire interval. Since \(f(x)\) will be positive whenever \(x\) is positive and both ends of the interval are positive, this criteria is satisfied.

  2. \(\int_{1}^{e^6}f(x)\,dx=1\)

We can test this criteria: \(\int_{1}^{e^6}f(x)\,dx=\int_{1}^{e^6}\frac{1}{6}x^{-1}\,dx={\displaystyle _{1}^{e^6}}[\frac{1}{6}\text{ln}(x)]=[(\frac{1}{6}\text{ln}(e^6))-(\frac{1}{6}\text{ln}(1))]=[(\frac{1}{6}(6))-(\frac{1}{6}(0))]=[1-0]=1\)

We can check this result in R:

f_x_7 <- function(x) {1 / (6 * x)}

as.numeric(integrate(f_x_7, lower = 1, upper = exp(6))[1])
## [1] 1

Therefore, \(f(x)\) satisfies both criteria for a pdf on the interval \([1,e^6]\).