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

\[ \int 4e^{-7x}dx \] \[ u=-7x \] \[ \frac {du}{dx}=-7\] \[ dx = - \frac {1}{7}du\]

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

\[ \int 4e^{-7x}dx = - \frac {4}{7} e^{-7x} + c\] 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.

\[\frac {dN}{dt} = - \frac {3150}{t^4} -220\] \[\int \frac {dN}{dt} = \int - \frac {3150}{t^4} -220 dt\] \[d(N)= \int - 3150t^{-4} -220 dt\] \[ N(t) = \frac{3150}{3t^{3}} - 220t + C \] At t=1, N(1) = 6350 \[ N(1) = 1050 - 220 + C = 6350\] \[ C = 5700\] \[ N(t) = \frac {1050}{t^{3}} - 220t + 5700\]

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

\[ \int_{4.5}^{8.5} (2x-9)dx \]

\[x^{2} - 9x |_{4.5}^{8.5} \] \[Area = ((8.5)^2-(9x8.5))-((4.5)^2-(9x4.5))\] \[ Area = 16 \]

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

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

eq1 <- function(x){
  return(x**2 -2*x - 2)
}
eq2 <- function(x){
  return(x + 2)
}
x <- seq(-2, 5, 1)
y <- eq2(x)

plot(eq1, from = -2, to = 5, line = "l", col = 'black')
lines(x,y, col = 'orange')

\[ y = x^2 - 2x -2 , y = x + 2\] \[ x^2 - 2x -2 = x + 2\] \[ x^2 - 3x -4 = 0, x=-1, x=4\] \[ Area = \int_{-1}^{4} x+2dx - \int_{-1}^{4} x^{2} -2x -2dx\] \[ = \int_{-1}^{4} -x^{2} +3x +4dx\]

\[ -\frac {x^{3}}{3} + 3\frac{x^{2}}{2} +4x |_{-1}^{4}\]

\[Area = -4^{3} /3+3(4^{2}) /2+4(4) - (-(-1)^{3} /3+3((-1)^{2}) /2+4(-1)\] \[ Area = 20.8333333\] 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.

Economic Order Quantity (EOQ) - It is a measurement used in the field of Operations, Logistics, and Supply Management. In essence, EOQ is a tool used to determine the volume and frequency of orders required to satisfy a given level of demand while minimizing the cost per order. (https://corporatefinanceinstitute.com/resources/knowledge/finance/what-is-eoq-formula/)

D: Annual Quantity Demanded Q: Volume per Order S: Ordering Cost (Fixed Cost) H: Holding Cost (Variable Cost)

\[Annual Ordering Cost = \frac{D}{Q} * S\] \[Annual Holding Cost = \frac{Q}{2} * H\] \[Annual Total Cost (TC) = \frac{D}{Q} * S + \frac{Q}{2} * H\] \[EOQ = \frac{dTC}{dQ} = \sqrt\frac{2SD}{H}\] \[= \sqrt\frac{2*8.25*110}{3.75} = 22\] 6. Use integration by parts to solve the integral below.

\[ \int ln (9x) * x^{6}dx \] \[ u = ln(9x), \frac{du}{dx} = \frac{1}{x}\] \[ \frac{dv}{dx} = x^{6}, v = \int x^{6}dx = \frac {1}{7}x^{7}\]

Using the formula (integration by parts):

\[ \int u \frac{dv}{dx}dx = uv - \int v \frac{du}{dv}dx\] \[ \int ln (9x) * x^{6}dx = ln(9x) . \frac {1}{7}x^{7} - \int \frac {1}{7}x^{7} . \frac{1}{x}dx \]

\[ = ln(9x) . \frac {1}{7}x^{7} - \int \frac {1}{7}x^{6}dx \]

\[ = ln(9x) . \frac {1}{7}x^{7} - \frac {1}{49}x^{7} + C \]

  1. Determine whether f(x) is a probability density function on the interval [1, e6] . If not, determine the value of the definite integral.

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

f(x) is a probability density function if the integral of f(x)=1

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

\[ = \frac{1}{6}*ln(e^{6}) - \frac{1}{6}*ln(1) = 1 \] The function is a PDF