```
knitr::opts_chunk$set(echo = TRUE)
library(tinytex)
```

Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/]

Use integration by substitution to solve the integral below.

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

\[u=-7x,\text{ }u'=-7x\ dx = -\frac{1}{7}\\ \int{4e^{-7x}}dx = 4 \int{e^u} du * u'\\ -\frac{4}{7}\int{e^u du}\\ -\frac{4}{7} e^u\\ -\frac{4}{7} e^{-7X}\]

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.

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

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

\[f(x) = 2x-9\\ area = \int_{4.5}^{8.5}{2x\ -\ 9\ dx}\\ area = (x^2 - 9x) \rvert_{4.5}^{8.5}\\ area = (8.5^2 - 9*8.5) - (4.5^2 - 9*4.5)\\ area = 16\]

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

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

The area between curves is the absolute value of the difference between their integrals.

\[area = \int{x^2 - 2x - 2\ dx} - \int{x + 2\ dx}\\ area = \int{x^2 - 2x - 2 - x -2\ dx}\\ area = \int{x^2 - 3x - 4\ dx}\\ area = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x + C\\\]

Find intersection points (since problem only asks for the bounded region). Use substitiution and solve for x:

\[x^2 - 2x - 2 = x + 2\\ x^2 - 3x -4 = 0\\ (x-4)(x+1) = 0\\ x = 4, -1\]

Evaluate the integral over -1 to 4:

\[area = (\frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x) \rvert_{-1}^{4}\\ area = (\frac{4^3}{3}-\frac{3*4^2}{2}-4*4) - (\frac{-1^3}{3}-\frac{3*-1^2}{2}-4*-1)\\ area = |-\frac{125}{6}| = \frac{125}{6}\]

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.

D = 110 (Demand, quantity to sell over year) Q = Volume per order S = 8.25 (Fixed ordering cost) C = Unit Cost H = 3.75 (Holding cost)

Ordering Cost:

\[\text{Number of Orders} = \frac{D}{Q}\]

Annual Ordering Cost:

\[\text{Annual Ordering Cost} = \frac{D}{Q} * S = 8.25\frac{110}{Q} = \frac{907.5}{Q}\]

Holding Cost:

\[\text{Annual Holding Cost} = \frac{Q}{2}*3.75 = 1.875Q\]

Total Cost with respect to Quantity:

\[\text{Annual Total Cost (T)} = \frac{907.5}{Q} + 1.875Q\]

Minimize Total Cost:

\[\frac{dT}{dQ} = \frac{907.5}{-2}*Q^{-2} + 1.875 = -473.75Q^{-2} + 1.875\\ \frac{-473.75}{Q^2}+1.875 = 0\\ Q^2 = \frac{473.75}{1.875} = \sqrt{252.667}\\ Q \approx 15.90\]

**Q = 16** to minimize overall storage cost

Use integration by parts to solve the integral below.

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

\[\int{f\ dg} = f*g - \int{g\ df}\\ \ \\ \ \\ f = ln(9x), df = \frac{1}{x}\ dx\\ dg = x^6\ dx , g = \frac{x^7}{7}\\ \ \\ \ \\ = ln(9x)*\frac{x^7}{7} - \int{\frac{x^7}{7}*\frac{1}{x}\ dx}\\ = ln(9x)*\frac{x^7}{7} - \frac{1}{7} \int{x^6}\ dx\\ = ln(9x)*\frac{x^7}{7} - \frac{1}{49} x^7 + C\\ = \frac{1}{49} x^7 (7*ln(9x) - 1) + C \]

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

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

Check 2 conditions:

- f(x) has to be non-negative - YES, we are told itβs over the interval \([1, e^6]\) which is positive at all points
- \(\int_{1}^{e^6}{f(x)\ dx} = 1\)

\[\int_{1}^{e^6}{\frac{1}{6x}\ dx} = \frac{1}{6} ln(x)\rvert_{1}^{e^6}\\ = \frac{1}{6} ln(e^6) - \frac{1}{6}ln(1)\\ = \frac{6}{6} - \frac{0}{6}\\ = 1\]

So, YES, \(f(x)\) meets both criteria for being a PDF.