Question 1

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

Solution

\[\int 4e^{-7x}dx\] \[u = -7x \\ du = -7dx \\ dx = \frac{du}{-7}\]

By pluggin the u and the du, the equation should be: \[\int 4e^{u}\frac{du}{-7}\] \[-\frac{4}{7}\int e^{-7x}du\] \[-\frac{4}{7}e^{-7x}+c\]

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.

Solution

\[\frac{dN}{dt} = -\frac{3150}{t^4} - 220\] \[N(t) = \int -\frac{3150}{t^4}dt - 220 dt \\ N(t) = \int -3150t^{-4} dt - 220 dt \\ N(t) = -3150(-4t^{-3}) - 220t + c \\ N(t) = \frac{12600}{t^3} - 220t + C \] By plugging in the value of 1 day and the level of bacteria contamination, the equation should be: \[6530 = N(t) = \frac{12600}{1^3} - 220(1) + C \] \[6530 = \frac{1}{12600} - 220 + C \\ 6750 = 12600 + C \\ C = -5850 \] So the final equation should be: \[N(t) = \frac{12600}{t^3} - 220t - 5850\]

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

Based on the graph of the total area, it is between 4.5 and 8.5 

area_function <- function(x){
  result = 2*x -9
  return (result)
}

min_int <- 4.5
max_int <- 8.5

total_area <- integrate(area_function, min_int, max_int)

total_area
## 16 with absolute error < 1.8e-13

Question 4

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

Solution

\[y=x^2-2x-2, y=x+2 \\ x^2-2x-2 = x+2 \\ 0 = x^2 - 3x - 4 \\ 0 = (x-4)(x+1)\] So x = 4 and -1

area_function2 <- function(x){
  result = x + 2
  return (result)
}

area_function3 <- function(x){
  result = x^2 -2*x - 2
  return (result)
}

min_int1 <- -1
max_int1 <- 4

total_area2 <- integrate(area_function2, min_int1, max_int1)
total_area3 <- integrate(area_function3, min_int1, max_int1)
area_under_curve <- round((total_area2$value - total_area3$value),5)

area_under_curve
## [1] 20.83333

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

\[f(x) = 8.25x + (3.75 \times \frac{110}{x}) \\ f(x) = 8.25x + (3.75 \times 110x^{-1}) \\ f(x) = 8.25x + 412.5x^{-1} \\ f(x) = 8.25 - 412.5x^{-2} \] By setting the value to 0 in order to minimize the cost, the equation should be: \[0 = 8.25 - 412.5x^{-2} \\ 8.25 = \frac{412.5}{x^2} \\ x^2 = \frac{412.5}{8.25} \\ x = \sqrt{\frac{412.5}{8.25}} \\ x = 7.071067812\] So the order is around 7. By plugging in the number, the equation should be: \[f(x) = 8.25 - 412.5x^{-2} \\ f(x) = 8.25 - 412.5(7)^{-2} \\ f(x) = 8.25 - \frac{412.5}{7^2} \\ f(x) = 8.25 - \frac{412.5}{49} \\ f(x) = 8.25 - 8.41 \\ f(x) = 1\]

Question 6

Use integration by parts to solve the integral below. \[\int \ln(9x)x^6dx\]

Solution

\[u = \ln(9x) \\ du = \frac{1}{x} \\ dv = x^6dx \\ v = \frac{x^7}{7}\] \[\int \ln(9x)x^6dx \\ \ln(9x) \times \frac{x^7}{7} - \int \frac{1}{x} \times \frac{x^7}{7} \\ \ln(9x) \times \frac{x^7}{7} - \int \frac{x^6}{7} \\ \frac{\ln(9x) \times x^7}{7} - \frac{x^7}{49} \\ \frac{7 \times \ln(9x) \times x^7}{49} - \frac{x^7}{49} \\ \frac{\ln(9x) \times 7x^7 - x^7}{49} \\ \frac{x^7(7\ln(9x) - 1)}{49} + c\]

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

Solution

\[f(x) = \frac{1}{6x} \\ f'(x) = \int_{1}^{e^6}\frac{1}{6x}dx \\ f'(x) = \frac{1}{6}\int_{1}^{e^6}\frac{1}{x}dx \\f'(x) = \frac{1}{6}\int_{1}^{e^6}\ln(x)dx \\ \frac{1}{6}(\ln(e^6)) - \frac{1}{6}(\ln(1)) = \frac{1}{6}(\ln(e^6)) - \frac{1}{6}(0) = \frac{1}{6}(\ln(e^6)) \\ \frac{1}{6}(6)(\ln(e)) = \ln(e) = 1\]

f(x) is determined to be the probability density function on the interval \[[1, e^6]\]