Question 1

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

Answer:
\(\int 4e^{-7x}dx\)
\(4\int e^{-7x}dx\)
\(4\int{e^u}du\) where \(u = -7x\)
\(4 e^u + C\)
\(4e^{-7x} + C\)

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.

\(\frac{dN}{dt} = -\frac{3150}{t^4}-220\)
\(N(t)=\frac{-3150}{3}t^{-3}-220t + C = 1050t^{-3}-220t + C\)
\(N(1) = 6530 = 1050 - 220 + C\)
\(C = 5700\)

So, the function \(N(t) = 1050t^{-3}-220t + 5700\).

Question 3

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

Answer:

From the graph, we know we can add up the rectangles for \(x = [5,8]\), and \(dx=1\):

\(\sum f(x)dx =\sum_{5}^{8} (2x-9) = 16\)

x <- 5:8
sum(2*x-9)
## [1] 16

Question 4

Find the area of the region bounded by the graphs of the given equations. \(y=x^2-2x-2, y=x+2\). From the plot below, we see the range is \([-1,4]\).

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

curve(y1, xlim=c(-3,5))

curve(y2, xlim=c(-3,5), add = TRUE)

To get the bound area, we can just calculate y2 - y1 = 20.8.

integrate(y2, lower = -1, upper = 4)$value - 
  integrate(y1, lower = -1, upper = 4)$value
## [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.

Question 6

Use integration by parts to solve the integral below.

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

Answer:

\(u = ln(9x)\)
\(du = \frac{9}{9x}dx = \frac{1}{x}dx\)

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

\(uv - \int{vdu} = ln(9x)\frac{x^7}{7} - \int{\frac{x^7}{7}\frac{1}{x}dx}\)
\(ln(9x)\frac{x^7}{7} - \int{\frac{x^6}{7}dx}\)
\(ln(9x)\frac{x^7}{7} - x^7 + C\)

Question 7

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

Answer:

A probability density function requires two properties:

  1. \(f(x) \ge 0\) for \([a,b]\).
  2. For the interval \([a,b]\), the total area must equal 1.

For 1), we can graph the interval and see all \(f(x) \ge 0\):

sample_function <- function (x) { 1/(6*x) }

curve(sample_function, xlim=c(1,exp(6)))

For 2), we can use the integrate function to verify the area = 1 for \([1,e^6]\):

integrate(sample_function, lower = 1, upper = exp(6))$value
## [1] 1