knitr::opts_chunk$set(echo = TRUE) library(tinytex) Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/] ## Problem 1 Use integration by substitution to solve the integral below. $\int{4e^{-7x}}dx$ $u=-7x,\text{ }u'=-7x\ dx = -\frac{1}{7}\\ \int{4e^{-7x}}dx = 4 \int{e^u} du * u'\\ -\frac{4}{7}\int{e^u du}\\ -\frac{4}{7} e^u\\ -\frac{4}{7} e^{-7X}$ ## 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. $\frac{dN}{dt} = \frac{-3150}{t^4} - 220\\ dn = (\frac{-3150}{t^4} - 220)\ dt\\ N(t) = \int{-3150*t^{-4} -220}\ dt\\ N(t) = \frac{-3150}{-3} * t^{-3} - 220t + C\\ N(t) = 1050 t^{-3} - 220 t + C\\ 6530 = N(1) = 1050 - 220 + C\\ C = 5700\\ N(t) = 1050 t^{-3} - 220 t + 5700$ ## Problem 3 Find the total area of the red rectangles in the figure below, where the equation of the line is $$f(x)=2x-9$$. $f(x) = 2x-9\\ area = \int_{4.5}^{8.5}{2x\ -\ 9\ dx}\\ area = (x^2 - 9x) \rvert_{4.5}^{8.5}\\ area = (8.5^2 - 9*8.5) - (4.5^2 - 9*4.5)\\ area = 16$ # Problem 4 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: $area = (\frac{1}{3}x^3 - \frac{3}{2}x^2 - 4x) \rvert_{-1}^{4}\\ area = (\frac{4^3}{3}-\frac{3*4^2}{2}-4*4) - (\frac{-1^3}{3}-\frac{3*-1^2}{2}-4*-1)\\ area = |-\frac{125}{6}| = \frac{125}{6}$ ## 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.

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

## Problem 6

Use integration by parts to solve the integral below.

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

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

## 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.

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

Check 2 conditions:

• f(x) has to be non-negative - YES, we are told it’s 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.