Problem 1:

Solve by substitution

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

Let \(z=-7x\). So \(dx=-\frac{1}{7}dz\)

\[ 4\int e^z \frac{-1}{7}dz=\\ \frac{-4}{7}\int e^zdz=\\ -\frac{4}{7}e^z+C=\\ -\frac{4}{7}e^{-7x}+C \]

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.

\[ \int\frac{dN}{dt}=\int\frac{-3150}{t^4}-220dt=\\ -\frac{12600}{t^3}-220t+C \]

Since \(N(1)=6530\), we can use this to find \(C\).

\[ N(1)=6530=-\frac{12600}{1^3}-220(1)+C\\ C=6290\\ N(t)=-\frac{12600}{t^3}-220t+6290 \]

Problem 3

The height of the rectangles is 1, 3, 5, and 7. Each is 1 wide. So the total area is 16.

(I was going to try to tell R to due numerical integration here with wide rectangles, but there is no documentation as to the method that the basic integrate method and pracma only specifies method for trapezoid method integrals.)

Problem 4

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

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

To find the intersections:

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

So the intersections occur at \(x=-1\) and \(x=4\). The area enclosed by these points has \(y=x+2\) above the other equations. So to find the area:

\[ \int_{-1}^4x+2dx-\int_{-1}^4x^2-2x-2dx=\\ \int_{-1}^{4}-x^2+3x+4dx=\\ -\frac{x^3}{3}+\frac{3x^2}{2}+4x\bigg|_{-1}^4=\\ -\frac{64}{3}+\frac{48}{2}+16-(\frac{-1}{3}+\frac{3}{2}-4)=\\ -\frac{63}{3}+\frac{45}{2}+20=\\ -21+22.5+20=\\ 21.5 \]

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.

Because this is a stepwise function, I was unable to create a nice function to optimize. Instead, I let R do the calculations. I made the assumption that flat irons sell at a constant rate. Also, instead of finding the derivative, given the ceiling function and the discrete values for orders, I just plugged in the integer values to find the solution.

orders<-function(x){ #x is the number of orders placed.
  cost<-x*8.25 #fixed cost per order
  storage<-0 #so I remember what's going on
  i=0
  while(ceiling((110/x)-i*110/365)>0){ #this is the total storage cost for a single shipment.  We want to have a delivery on the day we run out,
    #thus the break statement
    storage<-storage+(3.75/365)*ceiling((110/x-i*110/365))
    i<-i+1
    if(i==10000){
      break #just in case I wrote an infinite loop
    }
  }
  storage<-storage*x
  cost<-cost+storage
  return(cost)
}
orders(1)
## [1] 216.9144
orders(2)
## [1] 122.0342
orders(3)
## [1] 95.97945
orders(4)
## [1] 87
orders(5)
## [1] 84.91438
orders(6)
## [1] 86.36301
orders(7)
## [1] 89.68151

So there should be 5 orders of 22 irons each.

Problem 6:

Use integration by parts to solve the integral below.

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

Let \(u=\ln(9x)=\ln(9)+\ln(x)\). Let \(dv=x^6dx\). \(v=\frac{x^7}{7}\). \(du=\frac{1}{x}+\ln(9)xdx\).

So

\[ \int \ln(9x)x^6dx=(\frac{x^7}{7}\ln(9x))-\int\frac{x^7}{7}(\frac{1}{x}+\ln(9)x)dx=\\ (\frac{x^7}{7}\ln(9x))-\int\frac{x^6}{7}+\ln(9)\frac{x^8}{7}dx=\\ (\frac{x^7}{7}\ln(9x))-\frac{x^7}{56}-\ln(9)\frac{x^9}{63}+C \]

Given that we are not computing this numerically, I am leaving \(\ln(9)\) as a value instead of putting in an approximation.

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.

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

\[ \int_1^{e^6}\frac{1}{6x}dx=\\ \frac{1}{6}\int_1^{e^6}\frac{1}{x}dx=\\ \frac{1}{6}(\ln({e}^{6})-\ln(1))=\\ \frac{1}{6}(6\ln(e)-\ln(1))=\\ \frac{1}{6}(6\times1-0)=\\ 1 \]

Therefore, this is a probability distribution.