Question 1

Use integration by substitution to solve the integral below \(\int 4e^{-7x} dx\)

Solution

Let \(u = -7x\). Therfore, \[ \begin{aligned} du &= -7dx \\ dx &= \frac{-1}{7}du \end{aligned} \] Substituting \(u\) and \(dx\) into the integral \[ \begin{aligned} \int 4e^{-7x}dx &= \int 4e^{u}\begin{pmatrix}\frac{-1}{7}\end{pmatrix}du \\ &= \frac{-4}{7}\int e^{u}du \\ &= \frac{-4}{7}e^{u} + C \end{aligned} \] Substitute back \(x\) \[ \begin{aligned} \int 4e^{-7x}dx &= \frac{-4}{7}e^{-7x} + C \end{aligned} \]

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

Solution

\[ \begin{aligned} \frac{dN}{dt} &= \frac{-3150}{t^{4}} - 220 \\ N(t) &= \int\begin{pmatrix} \frac{-3150}{t^{4}} - 220 \end{pmatrix}dt \\ &= \int\begin{pmatrix} -3150t^{-4} - 220 \end{pmatrix}dt \\ &= -3150\frac{t^{-3}}{-3} - 220t + C \\ &= \frac{1050}{t^{3}} - 220t + C \\ \end{aligned} \] When \(t = 1\), \(N(1) = 6530\); \[ \begin{aligned} 6530 &= \frac{1050}{1^{3}} - 220*1 + C \\ &= 1050 - 220 + C \\ C &= 6530 - 1050 + 220 \\ &= 5700 \end{aligned} \] Therefore; \[ \begin{aligned} N(t) &= \frac{1050}{t^{3}} - 220t + 5700 \end{aligned} \]

Question 3

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

Solution

The red rectangles have a width of 1. We need to find the height at x = 5, 6, 7, and 8 \[ \begin{aligned} f(5) &= 2*5 - 9 = 1 \\ f(6) &= 2*6 - 9 = 3 \\ f(7) &= 2*7 - 9 = 5 \\ f(8) &= 2*8 - 9 = 7 \end{aligned} \] Note: The heights could also have been read off the graph Therefore the area of the red rectangles is; \[ \begin{aligned} Area &= 1*1 + 3*1 + 5*1 + 7*1 \\ &= 16 \end{aligned} \]

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

First find the points of interception; \[ \begin{aligned} x^2 - 2x - 2 &= x + 2 \\ x^2 - 2x - x - 2 - 2 &= 0 \\ x^2 - 3x - 4 &= 0 \\ (x - 4)(x + 1) &= 0 \\ \end{aligned} \] Therefore; \(x = 4\) and \(x = -1\) are the bounds of the two equations. Using \(x = 0\), we note that \(y = x + 2\) is above
\(y = x^2 - 2x - 2\)
Hence, to find the area of the region bounded by the two graphs we find the integral of the difference of \(y = x^2 - 2x - 2\)
from \(y = x + 2\) at the bounds \(x = 4\) and \(x = -1\). That is; \[ \begin{aligned} Area &= \int_{-1}^{4} (x + 2) - (x^2 - 2x - 2) dx \\ &= \int_{-1}^{4} -x^2 + 3x + 4 dx \\ &= \begin{bmatrix}-\frac{x^3}{3} + \frac{3x^2}{2} + 4x \end{bmatrix}_{-1}^{4} \\ &= \begin{pmatrix}-\frac{4^3}{3} + \frac{3*4^2}{2} + 4*4 \end{pmatrix} - \begin{pmatrix}-\frac{(-1)^3}{3} + \frac{3*(-1)^2}{2} + 4*(-1) \end{pmatrix} \\ &= 18.67 + 2.17 \\ &= 20.83 \end{aligned} \]

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.

Solution

Let lot size = N
Storage cost = \(\$3.75 * N\)
Fixed cost = \(\$8.25 * \frac{110}{N}\)
Total cost(TC) = storage cost + fixed cost = \(\$8.25 * \frac{110}{N} + \$3.75 * N\)
To find the lot size and number of order per year that will minimize inventory costs, differentiate TC with respect to N and set to 0: \[ \begin{aligned} \frac{\text{d}TC}{\text{d}N} = -\frac{907.5}{N^2} + 3.75 &= 0 \\ N^2 &= \frac{907.5}{3.75} \\ N &= \sqrt{242} \\ N &= 15.56 \approx 16 \\ \end{aligned} \] Optimum Lot size = 16, therefore
Optimum Number of orders = \(\frac{110}{16} = 6.88 \approx 7\)

Question 6

Use integration by parts to solve the integral \(\int (\ln{9x})*(x^6) dx\)

Solution

Let \[ \begin{aligned} u &= \ln(9x) \\ dv &= x^6dx \end{aligned} \] Then; \[ \begin{aligned} du &= \frac{1}{9x}*9 \\ &= \frac{1}{x} \\ v &= \frac{x^7}{7} \end{aligned} \] Substituting into; \[ \begin{aligned} \int u\:dv &= uv - \int v\:du \\ \int (\ln{9x})*(x^6)\:dx &= \frac{x^7\ln(9x)}{7} - \frac{1}{7}\int \frac{x^7}{x}\:dx \\ &= \frac{x^7\ln(9x)}{7} - \frac{1}{7}\int x^6\:dx \\ &= \frac{x^7\ln(9x)}{7} - \frac{x^7}{49} \\ \end{aligned} \] Therefore; \[ \begin{aligned} \int (\ln{9x})*(x^6)\:dx &= \frac{x^7\ln(9x)}{7} - \frac{x^7}{49} + C \\ \end{aligned} \]

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

Solution

Check that \(f(x) = \frac{1}{6x}\) meets the following conditions:
1. \(f(x) \geq 0\) for all \(x\) in the interval \([1, e^6]\)
2. The integral of \(f(x)\) over the interval \([1, e^6]\) is equal to 1

Is \(f(x) \geq 0\) for all \(x\) in the interval \([1, e^6]\)?

library(tidyverse)

# define function
f <- function(x){
  return (1/(6*x))
}

# define x value
x_values <- seq(1, exp(6), length.out = 1000)

# define corresponding y values
y_values <- f(x_values)

# create a data frame
df <- data.frame(x = x_values, y = y_values)

# plot
df %>% ggplot(aes(x = x, y = y)) +
  geom_line(color = 'blue') +
  labs(x = 'x', y = 'f(x)', title = "Plot of f(x) = 1/(6x) on interval [1, e^x]")

The plot of \(f(x) = \frac{1}{6x}\) is positive for all \(x \geq 0\). Therefore over the interval \([1, e^6]\), \(f(x)\) is greater than or equal to 0

Is the integral of \(f(x)\) over the interval \([1, e^6]\) is equal to 1? \[ \begin{aligned} \int_{1}^{e^6} \frac{1}{6x}\: dx &= \frac{1}{6}\int_{-1}^{e^6} \frac{1}{x} \: dx \\ &= \frac{1}{6} \begin{bmatrix}\ln x\end{bmatrix}_{1}^{e^6} \\ &= \frac{1}{6}({ln{e^6}} - ln{1}) \\ &= \frac{1}{6}(6 - 0) \\ &= 1 \end{aligned} \] Both conditions have been met. Therefore, \(f(x)\) is a probability density function over the interval \([1, e^6]\).