Question 1

Use integration by substitution to solve the integral below: \[\int4e^{-7x}dx\] Apply linearity: \[=4e^{-7}.\int x\;dx\] Now Solving \[\int x\;dx\] Apply power rule: \[\int x^{n}dx=\frac{x^{n+1}}{n+1}\;with\;n=1\\=\frac{x^2}{2}\] Plug in solved integrals: \[4e^{-7}. \int x\;dx\\=2e^{-7}x^2+C\]

Question 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.

Given \(t=1\) and \(N(1)=6530\)

\[\begin{aligned} \frac{d N}{d t} & =-\frac{3150}{t^4}-220=N(t)=-\int \frac{3150}{t^4} d t-200 \int d t+C \\ N(t) & =-3150 * t^{-4}-200 \int d t+C=-\frac{3150}{-3} t^{-3}-200(t)+C \end{aligned}\] Substitute value of \(t\) and \(N(t)\) \[\begin{gathered} N(1)=\frac{1050}{(1)^3}-220(1)+C \\ 6530=1050-220+C \\ C=6530-1050+220 \\ C=5700 \end{gathered}\] Substitute \(C\) value: \[N(t)=1050t^{-3}-200t+5700\]

Question 3

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

img

The above plot shows limit(lower = 4.5, upper = 8.5)

# function to find area
fun_area <- function(x) {
  return(2 * x - 9)
}

# use integrate to calc
(value <- integrate(fun_area, lower = 4.5, upper = 8.5))
## 16 with absolute error < 1.8e-13

Question 4

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

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

Let’s plot the lines:

# create function for each equation
eq_1 <- function(x) {
  return(x**2-2*x-2)
  }
eq_2 <- function(x) {
  return(x+2)
  }

# plot
plot(eq_1, -5, 5, col="red",ylab = "Y")
plot(eq_2, -5, 5, col="green",add=TRUE)

Lines intersect at -1 anbd 4.

Let’s find the area of intersection:

# define function to find intersection
fun_inter <- function(x) {
  return((x+2) - (x**2 - 2*x -2))
}

# use integrate to find area
(area <- integrate(fun_inter, lower = -1, upper = 4))
## 20.83333 with absolute error < 2.3e-13

Question 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.

Suppose half inventory keeps stock, then let \(x\) = size of iron

total cost = storage cost * x + oder cost / x

# set vars
stor_cost <- 3.75/2
order_cost <- 8.25*110

\[\begin{gathered} 1.875 * x^2-907.5=0 \\ 1.875 * x^2=907.5 \\ x^2=\frac{907.5}{1.875} \\ x=\sqrt{\frac{907.5}{1.875}} \end{gathered}\]

# find lot size
x <- sqrt(order_cost/stor_cost)

# amount of times can order
times <- 110/x

cat("Each year irons can be ordered",times,"times in batches of",x)
## Each year irons can be ordered 5 times in batches of 22

Question 6

Use integration by parts to solve the integral below.

\[\int \ln (9 x) \cdot x^6 d x\]

Let’s use the integration by parts formula:

\[\int u v d x=u \int v d x-\int u^{\prime}\left(\int v d x\right) d x\]

Where:

\[\begin{aligned} & \mathrm{u}=\ln (9 x) \\ & \mathrm{du}=\frac{1}{x} d x \\ & v=x^{\wedge} 6 \\ & d v=\frac{1}{7} x^7 \end{aligned}\]

\[\begin{aligned} & \int(u v) d x=\ln (9 x) * \frac{1}{7} x^7-\int \frac{1}{x} \frac{1}{7} x^7 d x \\ & \int(u v) d x=\ln (9 x) * \frac{1}{7} x^7-\frac{1}{7} \int x^6 d x \\ & \int(u v) d x=\ln (9 x) * \frac{1}{7} x^7-\frac{1}{49} x^7+C \end{aligned}\]

Question 7

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

\[f(x) = \frac{1}{6x}\] \[\int_1^{e^6} \frac{1}{6 x} d x=\left.\frac{1}{6} * \ln x\right|_1 ^{e^6}=\frac{1}{6} \ln \left(e^6\right)-\frac{1}{6} \ln (1)=\frac{1}{6}[6-0]=1\]