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

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

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

\[ \frac{dN}{dt}=-\frac{3150}{t^4}-220\\ N(t)= \int -\frac{3150}{t^4}-220 \, dt\\ N(t)= \int -3150t^{-4}-220 \, dt\\ N(t)=-3150(\frac{1}{-3})t^{-3}-220t+C\\ N(t)=\frac{3150}{3t^3}-220t+C\\ 6530=\frac{3150}{3(1)^3}-220(1)+C\\ 6530=1050-220+C\\ 6530=830+C\\ 5700=C\\ N(t)=\frac{1050}{t^3}-220t+5700\\ \]

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x-9\).
library(png)
img <- readPNG("/Users/mollysiebecker/Desktop/Screenshot 2024-04-21 at 11.36.46 PM.png")
plot(1:2, type="n", ann=FALSE, axes=FALSE)
rasterImage(img, 1, 1, 2, 2)

\[ \Delta x=1 \\ n=4\\ A = \sum_{i=1}^{n} f(x_i) \cdot \Delta x\\ A = \sum_{i=1}^{4} f(x_i) \cdot 1\\ A = 2(5)-9+2(6)-9+2(7)-9+2(8)-9\\ A = 2(26)-36\\ A = 16\\ \]

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

\[ y=x^2-2x-2, y=x+2\\ x^2-2x-2=x+2\\ x^2-3x-4=0\\ x={4,-1}\\ A=\int_{-1}^{4}(x+2)-(x^2-2x-2)\,dx\\ A=\int_{-1}^{4}-x^2+3x+4\,dx\\ A=\frac{-1}{3}(4)^3+\frac{3}{2}(4)^2+4(4)-(\frac{-1}{3}(-1)^3+\frac{3}{2}(-1)^2+4(-1))\\ A=-\frac{64}{3}+24+16-(\frac{1}{3}+\frac{3}{2}-4)\\ A=-\frac{64}{3}+24+16-\frac{1}{3}-\frac{3}{2}+4\\ A=-\frac{65}{3}-\frac{3}{2}+44\\ A=-\frac{130}{6}-\frac{9}{6}+\frac{264}{6}\\ A=\frac{125}{6} \approx 20.83\\ \]

  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.

The number of orders \(n\) is the expected demand divided by the lot size \(L\), and the amount of inventory at a given time can be calculated as half of the lot size, if we assume the inventory decreases linearly. \[ C=8.25(\frac{110}{L})+3.75(\frac{L}{2})\\ C'=-907.5L^{-2}+1.875\\ 0=-907.5L^{-2}+1.875\\ \frac{1.875}{907.5}=L^{-2}\\ L= \sqrt{\frac{907.5}{1.875}}\\ L=22\\ n=\frac{110}{22}\\ n=5\\ \] A lot size of 22, with 5 orders per year, will minimize the expected costs.

  1. Use integration by parts to solve the integral below. \[ \int ln(9x) \cdot x^6 \, dx\\ u=ln(9x)\\ dv=x^6\\ \int udv=uv-\int vdu\\ \int ln(9x) \cdot x^6=ln(9x) \cdot \frac{1}{7}x^7-\int \frac{1}{7}x^7 \cdot \frac{1}{x} dx\\ =ln(9x) \cdot \frac{1}{7}x^7-\frac{1}{49}x^7+C\\ =\frac{7ln(9x)-1}{49}(x^7)+C\\ \]

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

\[ f(x)=\frac{1}{6x}\\ \int_1^{e^6} \frac{1}{6x} \ dx\\ =\frac{1}{6}ln(e^6)-\frac{1}{6}ln(1)\\ =1-0\\ =1 \] Yes, \(f(x)\) is a probability density function on the given interval because it is non-negative, and the integral equals 1.