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

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

Let \(u = -7x\)

\(\frac{du}{dx} = -7\)

\(dx = - \frac{1}{7} du\)

Now, let’s solve for the given problem scenario

\(\int e^{-7x}dx\) = \(-\frac{1}{7} \int e^u du\)

=\(-\frac{e^u}{7}\)

Now plugging \(u=-7x\)

\(= -\frac{1}{7}e^{-7x} + c\)

  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 6350 bacteria per cubic centimer.

\(\frac{dN}{dt} = - \frac{3150}{t^4}-220\)

\(=\int \frac{dN}{dt} = \int - \frac{3150}{t^4}-220 dt\)

\(= \int -3150t^{-4}-220dt\)

\(N(t) = \frac{-3150}{-3}t^{-3} - 220t + c = \frac{1050}{t^3} - 220t + c\)

\(N(1) = 6350 = \frac{1050}{1^3} - 220(1) + c = 1050 - 220 + c = 830 + c\)

Now solving for c, \(6350 - 830=c\), c = 5520. Therefore, solution for \(N(t)\) is: \[N(t) = \frac{1050}{t^3} - 220t + 5520\]

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

per graph above, the definite integrate limit for the function will be from values 4.5 to 8.5

\(F(x) = \int_{4.5}^{8.5} f(x) dx\)

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

\(= x^2-9x + c\)

Given that the boundaries are from 4.5 to 8.5, we can perform \(F(8.5) - F(4.5)\)

\(F(8.5) - F(4.5)\)

\(= 8.5^2 - 9*8.5 - (4.5^2 - 9*4.5)\)

\(-4.25-(-20.25)\)

\(= 16\)

So, the exact area under the curve is: 16.

  1. Find the area of the region bounded by the graphs of the given equations \(y = x^2 - 2x - 2, \space y = x + 2\)
#Create the funcions
fn1 <- function(x){
  return(x**2 -2*x - 2)
}

fn2 <- function(x){
  return(x + 2)
}

x <- seq(-2, 5, 1)
y <- fn2(x)

# Plot the graphs
plot(fn1, from = -2, to = 5, line = "l", col = 'red') # function 1 
lines(x,y, col = 'blue') # function 2

library(rootSolve)
## Warning: package 'rootSolve' was built under R version 3.4.4
v_coefficient <- c(-4, -3, 1)
ans <- polyroot(v_coefficient)
print(paste0("The equations equal to each other at: ", round(ans,3)))
## [1] "The equations equal to each other at: -1+0i"
## [2] "The equations equal to each other at: 4+0i"

To get the area between the two curves \(\int_{-1}^{4} x + 2 dx - \int_{-1}^{4} x^2-2x-2dx\)

Eq 1
\(\int_{-1}^{4} x + 2 dx = \frac{x^2}{2} + 2x + C\)

\(=\frac{16}{2} + 2*4 - (\frac{1}{2} +2*-1) = 8 + 8 - (\frac{1}{2} - 2) = 17.5\)

Eq 2
\(\int_{-1}^{4} x^2-2x-2dx = \frac{x^3}{3} - x^2 - 2x + C\)

\(= \frac{64}{3} - 16 - 2*4 - (\frac{-1}{3} - 1 - 2*-1) = -10/3\)

Now, plugging values \(17.5 - \frac{-10}{3}\)

\(=20.83333\)

integrand <- function(x){x + 2 - (x**2 -2*x - 2)}
integrate(integrand, -1, 4)
## 20.83333 with absolute error < 2.3e-13
  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 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.

EOQ stands for Economic Order Quantity. 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 \((TC) = \frac{D}{Q} S + \frac {Q}{2}H\)

\(EOQ = \frac{dTC}{dQ} = \sqrt{\frac{2SD}{H}}\)

S <- 8.25
H <- 3.75
D <- 110 

EOQ <- function(S,H,D){
  return(sqrt(2*D*S/H))
}

print(paste0("Number of orders per year that will minimize inventory costs: ", EOQ(S,H,D)))
## [1] "Number of orders per year that will minimize inventory costs: 22"
  1. Use integration by parts to solve the integral below:

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

Applying formula for integration by parts : \(\int f(x) * g'(x) dx = f(x)g(x) - \int f'(x)g(x)dx\)

\(f(x) = ln(9x)\) and \(g'(x) = x^6\)

\(f'(x) = \frac{9}{9x} = \frac{1}{x}\) and \(g(x) = \frac{x^7}{7}\)

Now Solving for \(\int ln(9x) * x^6 dx\)

\(=ln(9x) \frac{x^7}{7} - \int \frac{1}{x} *\frac{x^7}{7} dx\)

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

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

  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 \(f(x)=1\)

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

\(= \frac{1}{6} * ln(e^6) - \frac{1}{6}*ln(1) = 1\)

Hence, this function is a probability density function.