• Question 1
  • Question 2
  • Question 3
  • Question 4
    • Solution 1
  • Question 5
  • Question 6
  • Question 7

Question 1

Use integration by substitution to solve the integral below.

\[\int 4e^{-7x} dx\] Let \(u = -7x\)

\(\frac{du}{dx}=-7\) —> Solving for \(dx\).

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

\(\int 4e^{-7x} dx\)

\(=\int 4e^{u} \cdot \left (-\frac{1}{7}du \right )\)

\(= -\frac{4}{7}\int e^{u} du\)

\(= -\frac{4}{7} e^{u}\) where \(u = -7x\)

\(= -\frac{4}{7} e^{-7x} + C\)

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.

\(\frac{dN}{dt}= N'(t)=-\frac{3150}{t^{4}}-220\)

\(N(t)=\int N'(t) dt = \int \left (-\frac{3150}{t^{4}}-220\right ) dt\)

\(N(t)=-\int \frac{3150}{t^{4}}dt-\int220dt\)

\(N(t)=-3150 \int \frac{1}{t^4}dt - 220t\)

\(N(t)=-3150 \cdot -\frac{1}{3t^3} - 220t + C\)

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

Now we find a function \(N( t )\) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

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

\(N(1) = \frac{1050}{1^{3}}- 220(1) + C = 6530\)

\(830+C = 6530\)

\(C = 6530 -830 = 5700\)

$N( t ) = - 220t + 5700 $

Question 3

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

knitr::include_graphics("~/Desktop/Question3.png")

Find the area under the line between x = 4.5 and x = 8.5

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

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

\(=x^2-9x|_{4.5}^{8.5}\)

\(=[(8.5)^2-9(8.5)]-[(4.5)^2-9(4.5)]\)

\(Area = 16 \; square \; units\)

\(Area_{rectangle}= length\cdot width\)

\(Area_{1}=(1-0)(5.5-4.5)=1\)

\(Area_{2}=(3-0)(6.5-5.5)=3\)

\(Area_{3}=(5-0)(7.5-6.5)=5\)

\(Area_{4}=(7-0)(8.5-7.5)=7\)

\(Total \;Area=Area_{1}+Area_{2}+Area_{3}+Area_{4} = 1+3+5+7=16 \; square \; units\)

Question 4

Find the area of the region bounded by the graphs of the given equations. \[y=x^2-2x-2,y=x+2\]

Solution 1

library(ggplot2)

x_values <- seq(-5,5, by = 0.1)

y1 <- function(x) x^2 - 2*x -2
y2 <- function(x) x + 2

data <- data.frame(x = x_values, y1 = y1(x_values), y2 = y2(x_values))

ggplot(data, aes(x = x)) + 
  geom_line(aes(y = y1), color = "blue", size = 1, linetype = "solid") + 
  geom_line(aes(y = y2), color = "green", size = 1, linetype = "solid") + theme_minimal()
## Warning: Using `size` aesthetic for lines was deprecated in ggplot2 3.4.0.
## ℹ Please use `linewidth` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.

Find the point of intersection(s) by equating the two equations together and solve for x.

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

\(x^2-3x-4=0\)

\((x-4)(x+1)=0\)

\(x=4\) or \(x=-1\)

Formula for the area between two curves: \[Area = \int_{a}^{b}[f(x)-g(x)]dx \]

\(Area = \int_{-1}^{4}[(x+2)-(x^2-2x-2)]dx\)

\(Area = \int_{-1}^{4}[-x^2+3x+4]dx\)

\(Area = [-\frac{x^3}{3}+3\cdot\frac{x^2}{2}+4x]_{-1}^{4}\)

\(Area = [-\frac{4^3}{3}+\frac{3\cdot 4^2}{2}+4(4)]-[-\frac{(-1)^3}{3}+\frac{3\cdot (-1)^2}{2}+4(-1)]\)

\(Area = -\frac{64}{3}+\frac{48}{2}+16-\frac{1}{3}-\frac{3}{2}+4\)

\(Area = -\frac{65}{3} + \frac{45}{2} + 20\)

\(Area = -\frac{130}{6}+\frac{135}{6}+\frac{120}{6}=\frac{125}{6}\)

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.

Let \(x\) be the number of flat irons per order. Thus, the number of orders is \(\frac{110}{x}\).

The yearly storage cost is the storage cost per iron times the average number of irons. \(\$3.75 \cdot \frac{x}{2}\)

The yearly order cost is the cost fixed cost per order times the number of orders \(\$8.25\cdot \left (\frac{110}{x} \right)\)

The inventory cost is \(f(x)=\$3.75 \cdot \frac{x}{2} + \$8.25\cdot \left (\frac{110}{x} \right) = 1.875x+\frac{907.5}{x}\)

Find the lot size and the number of orders per year that will minimize inventory costs:

\(f'(x)= 1.875-\frac{907.5}{x^2}=0\)

\(1.875=\frac{907.5}{x^2}\)

\(1.875x^2=907.5\)

\(x^2=\frac{907.5}{1.875}\)

\(x= \sqrt{484}\)

\(x = 22\) There should be 22 flat irons per order.

The number of orders is \(\frac{110}{x}=\frac{110}{22}=5\).

Question 6

Use integration by parts to solve the integral below.

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

Intregral by parts: \(\int fg'=fg-\int f'g\)

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

To find the derivative of \(f = ln(9x)\), we need apply the differential rule: \([ln(u(x)]'=\frac{1}{u(x)}\cdot u'(x)\)

Then, \(f'=\frac{1}{9x}\cdot 9=\frac{1}{x}\)

\(g = \int g' = \int x^6 dx = \frac{x^7}{7}\)

\(\int fg'=fg-\int f'g\)

\(\int ln(9x)\cdot x^6 dx\)

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

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

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

Question 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.

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

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

\(=\frac{1}{6}ln(x)|_{1}^{e^6}\)

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

\(=\frac{1}{6}\cdot 6-\frac{1}{6}\cdot 0\)

\(= 1\)