1. Use integration by substitution to solve the integral below

\[\int 4e^{-7x}dx\]

let \(u = -7x\)

then \(du = -7dx\)

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

Substituting into the original equation:

\[4\int e^{u}*(-\frac{1}{7})du\]

\[-\frac{4}{7}\int e^{u}du\] \[-\frac{4}{7}e^{u} + C\]

Substituting the value of u back in:

\[-\frac{4}{7}e^{-7x} + C\]

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.

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

\[N(t) = 1050t^{-3}-220t + C\]

Given that \(N(1) = 6530\), we can solve for C:

\[6530 = 1050-220 + C\]

\[C = 6530 + 220 - 1050\] \[C = 5700\]

Therefore:

\[N(t) = 1050t^{-3} - 220t + 5700\]

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

The area of the rectangles is:

\[f(5) + f(6) + f(7) + f(8)=\] \[(2*5-9)+(2*6-9)+(2*7-9)+(2*8-9)=\] \[(10-9)+(12-9)+(14-9)+(16-9)=\] \[1+3+5+7=\] \[16\]

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

\[y=x^{2}-2x-2\]

\[y = x+2\]

To start, let us find the points of intersection:

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

To confirm, let’s plot these two lines:

curve(x^2 -2*x-2, lwd = 2, xlim=c(-2, 5),col = "blue")
curve(x+2, lwd = 2, xlim=c(-2, 5), add = TRUE, col = "green")

To calculate the are between the two we need to calculate the following:

\[A=\int_{-1}^{4}(Top - Bottom)dx\]

From the graph we see that \((x+2)\) lies above \((x^{2}-2x-2)\) in this region

\[A = \int_{-1}^{4}((x+2)-(x^{2}-2x-2))dx\] \[A = \int_{-1}^{4}-x^{2}+3x+4 dx\] \[A =(-\frac{1}{3}x^{3}+\frac{3}{2}x^2+4x) \Big|_{-1}^{4}\] \[=(-\frac{1}{3}4^{3}+\frac{3}{2}4^2+4(4))-(-\frac{1}{3}(-1)^{3}+\frac{3}{2}(-1)^2+4(-1))\] \[=(-\frac{64}{3}+\frac{3*16}{2}+16)-(\frac{1}{3}+\frac{3}{2}-4)\] \[=-\frac{65}{3}+\frac{45}{2}+20\] \[A = 20.83\]

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.

library(Deriv)
## Warning: package 'Deriv' was built under R version 3.6.3
#Let us set x as the number of orders per year, then:
#Lot Size <- 110/x
#Cost to store <- 3.75*(110/x)
#Order cost per year <- x * 8.25


cost <- function(x){
  3.75*110/x + 8.25*x
}
#To find the lowest value we must differentiate and solve at 0:
derivCost <- Deriv(cost)
derivCost
## function (x) 
## 8.25 - 412.5/x^2

\[8.25-\frac{412.5}{x^{2}}=0\]

\[\frac{412.5}{x^{2}}=8.25\] \[412.5=8.5x^{2}\] \[x = 7.07\]

Rounding to the nearest integer gives us 7 orders, with 16 units in each (with 14 in the last).

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

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

\(u = ln(9x)\)

\(du = \frac{1}{x}dx\)

\(dv = x^{6}dx\)

\(v = \int x^{6} dx = \frac{1}{7}x^{7}\)

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

\[ =\frac{ln(9x)*x^{7}}{7} - \frac{x^{8}}{7*8}+C\] \[=\frac{8ln(9x)x^{7}-x^{8}}{56}+C\]

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.

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

The criteria for a PDF are:

  1. \(f(x) \ge 0\), for all x
  2. \(\int_{-\infty}^{+\infty}f(x)dx = 1\).

This \(f(x)\) does not meet criteria 1, because for negative values of x, \(f(x) < 0\).

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

\[F(x) =\frac{1}{6}ln(x) \Big|_{1}^{e^{6}}\] \[F(x) = \frac{1}{6} * (ln(e^{6})-ln(1))\] \[F(x) = \frac{1}{6} * (6-0)\] \[F(x) = 1\]

The \(f(x)\) gven meets the second criteria of being a PDF, but due to it not meeting the first, it is not a PDF.