Question 1

Use integration by substitution to solve the integral below.

\[\int 4e^{-7x}~dx\]

# substitution with u leads the following equation
u = -7x

du = -7dx

dx=du/-7

\[\int 4e^{u}~\frac{du}{-7}\] \[\frac{-4}{7}\int e^{u}~du\]

\[\frac{-4}{7}e^{u} + C\] \[\frac{-4}{7}e^{-7x} + C\]

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.

\[N = \int -\frac{3150}{t^{4}} - 220~dt\] \[N = \int -\frac{3150}{t^{4}}dt - \int220~dt\] \[N = -3150\int {t^{-4}}dt - \int220~dt\] \[N = \frac{3150}{3}t^{-3} - 220t + C\] Solve for C

C <- 6530 - (3150/3)*(1**-3) + 220*1
C
contamination <- function(t){
  val = (3150/3) * (t**-3) - 220*t + 5700
  
  return(val)
}

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.

\[\int_{4.5}^{8.5}2x-9 = (8.5^{2}-9*8.5) - (4.5^{2}-9*4.5) = 16\]

Question 4

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

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

f <- function(x) {x**2 - 3*x - 4}

integrate(f, 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

C = cost x = number of orders n = lot size

therefore we know that \[x*n = 110\]

and \[x* = 110/n\] and finally we can construct the equation:

\[C = 8.25n + 3.75*\frac{110}{n} = 8.25n + \frac{375}{n}\] Solve for derivative

\[C' = 8.25 - \frac{375}{n^2}\] Set derivative to 0 and solve for n

\[8.25 - \frac{375}{n^2} = 0\] \[8.25 = \frac{375}{n^2}\]

\[n^2 = \frac{375}{8.25} = 45.4545\]

n <- sqrt(375/8.25)
n
## [1] 6.741999

Order no more than 7 orders per year.

Question 6

Use integration by parts to solve the integral below.

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

u = ln(9x)

dv/dx = x^6

du = (1/x)dx

dv = x^6dx

v = (1/7)x^7

putting it all together

\[ln(9x)\frac{1}{7}x^7 - \int \frac{1}{7}x^7\frac{1}{x}~dx = ln(9x)\frac{1}{7}x^7 - \frac{1}{7}\int x^6~dx\] \[ln(9x)\frac{1}{7}x^7 - \frac{1}{49}x^7 + C\] # 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.

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

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

integrate(f, lower=1, upper = exp(6))
## 1 with absolute error < 9.3e-05