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

\[\int4e^{-7x}dx\]

\[u = -7x\] \[du=-7dx\] \[-\frac{du}{7}=dx\]

\[\int4e^{-7x}dx=-\frac{4}{7}\int e^udu\] \[=-\frac{4}{7}e^u+C=-\frac{4}{7}e^{-7x}+C\]

  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.

\[\frac{dN}{dt}=\frac{3150}{t^4}-220\] \[N(t)=\int (-\frac{3150}{t^4}-220)dt = \int -3150t^{-4}dt-\int220 dt\] \[=\frac{3150t^{-3}}{3}-220t+C\] \[N(1) = 6530 = \frac{3150}{3(1)^3}-220(1)+C \] \[C = 5700\] \[N(t) =\frac{3150}{3(t)^3}-220(t)+5700 \]

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

\[Area = \int_{4.5}^{8.5}(2x-9)dx=\left [ \frac{2x^2}{2}-9x\right]_{4.5}^{8.5}\] \[=(8.5^2-9(8.5))-(4.5^2-9(4.5))=16\]

  1. Find the area of the region bounded by the graphs of the given equations. \[y = x^2-2x-2,y = x+2\]
curve(x^2 - 2*x - 2, -5, 7, col = "blue")
curve(x + 2, -5, 7, add=T, col="red")

f1 <- function(x) x^2-2*x-2
f2 <- function(x) x+2
area = integrate(f2, lower = -1,upper = 4)$value -integrate(f1,lower = -1, upper = 4)$value
print(paste0('The area shading the intersection of these two functions is ',round(area,3)))
## [1] "The area shading the intersection of these two functions is 20.833"
  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.

N = number of orders/year L = Lot size

We want number of orders x lot size = 110 N*L = 110

Total costs = ordering cost + holding cost ordering cost = 8.25*N

Assuming the inventory is used up uniformly throughout the year, the average inventory count will be L/2. So on average the there will be L/2 in the inventory per year.

holding cost = 3.75(L/2) = 3.75(110/2N)

The total cost equation: \[Total Cost = 8.25N+3.75\frac{110}{2N}\]

In order to minimize costs, the rate of total cost with respect to L, lot size, is looked at: \[\frac{dT}{dN} = 8.25-\frac{206.25}{N^2}=0\]

When dT/dN = 0, the slope is 0 and that is the lowest cost per number of orders \[8.25=\frac{206.25}{N^2}\] \[N = 5\]

\[N*L=110\] \[5*L=110\] \[L = 22\]

Therefore there needs to be 5 orders of 22 flat irons to minimize total costs.

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

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

\[\int ln(9x)x^6dx=ln(9x)\frac{x^7}{7}-\int \frac{1}{x}\frac{x^7}{7}dx\] \[=ln(9x)\frac{x^7}{7}-\int \frac{x^6}{7}dx=ln(9x)\frac{x^7}{7}-\frac{1}{7}\frac{x^7}{7}+C\] \[\frac{7x^7ln(9x)-x^7}{49}+C\]

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

In order for a function to be a probability density function the function must be non-negative on the interval. This is because a probability cannot be negative. Additionally, the sum under the curve must be 1. The total probability of all probabilities must be 100%. \[\int_{1}^{e^6}\frac{1}{6x}=\frac{1}{6}ln(x)|_1^{e^6}\] \[=\frac{1}{6}ln(e^6)-\frac{1}{6}ln(1)=\frac{1}{6}(6)-0=1\]

From the interval 1 to e^6, f(x) will be positive. The integral of the function at this interval is 1. All conditions have been met and f(x) is a PDF.