ASSIGNMENT 13

Problem 1

Use integration by substitution to solve the integral below.

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

Answer 1

Let \(u=-7x\), then \(du = -7dx\).

\[\begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} \\ &= \int{\frac{-4}{7}e^u du} \\ &= \frac{-4}{7}e^u+constant \\ &= -\frac{4}{7}e^{-7x}+ constant \end{split}\]

Problem 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.

Answer 2

\(N(1)= 6530\)

\(\frac{dN}{dt} = -\frac{3150}{t^4} - 220\)

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

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

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

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

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

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

\(6530 = 1050 - 220 + C\)

$C = 6530 - 1050 + 220 = 5700 $

\(N(t) = -\frac{1050}{t^3} - 220t + 5700\)

Problem 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 3

f = function(x){
  (2*x) - 9
}
integrate(f,4.5,8.5)
## 16 with absolute error < 1.8e-13

Problem 4

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

\[y = x^2 - 2x - 2\]

\[y = x + 2\]

Answer 4

Using R functions to find the difference between areas under the curve

library(cubature)

f1 = function(x){
  (x^2)- (2*x) -2
}

f2 = function(x){
  x+2
}

ggplot(data=data.frame(x=0), mapping=aes(x=x)) +
  stat_function(fun = f1, color='red') +
  stat_function(fun = f2, color='blue') +
  xlim(-5, 5) + 
  scale_colour_manual(name = 'color', values =c('blue'='blue','red'='red'), labels = c('y=x+2','y=x^2-2x-2'))

#  legend("top", c("y=x^2-2x-2", "y=x+2"), fill=c("red", "blue"))

paste0("Area of the region bounded by the above graphs of the given equations:",hcubature(f2,-1,4)$integral - hcubature(f1,-1,4)$integral)
## [1] "Area of the region bounded by the above graphs of the given equations:20.8333333333333"

Problem 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.

Answer 5

  • Number of orders per year = n
  • Lot size = s
  • Cost = c

n*s = 110

s = 110/n

Holding costs are calculated by multiplying the per-unit annual holding cost by the average level of inventory. The average inventory level is equal to the number of ordered items, divided by two.

The holding costs using the formula described above would be 3.75 * (s/2).

\(c = 8.25n + \frac{3.75s}{2}\)

c = 8.25n + 1.875s

c = 8.25n + 1.875*100/n

\(c'\) = 8.25 - 206.25/\(n^2\)

\(c'\) = 0

n = sqrt(206.25/8.25)

n = 5

s = 110/n = 22

Hence, 5 orders per year for a lot size of 22.

Problem 6

Use integration by parts to solve the integral below.

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

Answer 6

Let \(u= ln(9x)\)

\(\frac{du}{dx}=\frac{1}{x}\)

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

\(v = \int{x^6 dx} = \frac{1}{7}x^7\)

\(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)

\[\begin{split} \int{ln(9x) \times x^6 dx} &= \frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^7 \times \frac{1}{x} dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^6 dx} \\ &=\frac{7}{49}x^7 \times ln(9x) - \frac{1}{49}x^7 + constant \\ &=\frac{1}{49}x^7 (7ln(9x) - 1) + constant \\ \end{split}\]

Problem 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.

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

Answer 7

\(F(x)=\int_1^{e^6}{f(x)dx}=1\) where \(f(x) = \frac{1}{6x}\)

\(F(x) = \int{_1^{e^6}\frac{1}{6x}dx}\)

\(= \frac{1}{6}\int_1^{e^6}{\frac{1}{x}dx}\)

\(= \frac{1}{6}ln(x)|_1^{e^6}\)

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

\(= \frac{1}{6}[6-0]\)

\(F(x) = 1\)

At interval \([1, e^6]\), the integral function F(x) resolves to 1. Therefore, F(x) is a probability density function at this interval.