Data 605 Homework Wk13

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library( ggplot2 )

#1

Use integration by substitution to solve the integral below. \[\int 4e^{-7x}dx\] 1. move constant in front \[\longrightarrow 4 \int e^{-7x}dx\] 2. substitute in \(u=-7x\) \[-4\int e^udu\longrightarrow -4\int -\frac{1}{7}e^udu\longrightarrow -\frac{4}{7}\int e^u du\] 3. apply common form for \(\int e^udu = e^u\) \[\longrightarrow -\frac{4}{7}e^u\] 4. substitute back in for u \[\longrightarrow -\frac{4}{7}e^{-7x}\] Therefore, we find that \[\int 4e^{-7x}dx \longrightarrow -\frac{4}{7}e^{-7x}+C\]

#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.

Find \(N(t)\) given \(\frac{dN}{dt} = -\frac{3150}{t^4}-220\):

\[\frac{dN}{dt} = -\frac{3150}{t^4}-220 \longrightarrow \] take the integral of both sides of the equation to find N(t) \[\int \frac{dN}{dt} = \int \left( -\frac{3150}{t^4}-220 \right)dx \longrightarrow \] \[\int \frac{dN}{dt} = \int \left( -3150t^{-4}-220 \right)dx \longrightarrow \] \[N(t) = -\frac{3150 \cdot t^{-3}}{-3}-220t + C \longrightarrow \] \[N(t) = -\frac{3150 \cdot t^{-3}}{-3}-220t + C \longrightarrow \] \[N(t) = -1050 \cdot t^{-3}-220t + C \longrightarrow \] Solve for \(C\) given the observation that at \(t=1\), $N(t)=6530: \[6530 = -1050 \cdot (1)^{-3}-220\cdot (1) + C \longrightarrow \] \[6530 = 830 + C \longrightarrow C = 5700\]

#3

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

Area = 16

#4

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

Start by visualizing the 2 functions

p <- ggplot(data = data.frame(x = 0), mapping = aes(x = x))
fun.1 <- function(x) x^2 - 2*x -2
fun.2 <- function(x) x +2
p + stat_function(fun = fun.1) + stat_function(fun = fun.2) + xlim(-2,5)

great, so we would like to find: \[Area = \int_{-1}^4 [(x + 2) - (x^2 -2x-2)]dx \rightarrow \int_{-1}^4 [-x^2 + 3x +4 ]dx \rightarrow\] find the difference between this expression evaluated at 4 and -1: \[\frac{-x^3}{3} + \frac{3x^2}{2} + 4x\]

#ealuated x=4
x <- 4
val4 <- -x^3/3 + 3*x^2/2 + 4*x
x <- -1
valneg1 <- -x^3/3 + 3*x^2/2 + 4*x
Area <- val4 - valneg1
Area

## [1] 20.83333

#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.

Consider the sales for a year: \[110 \mbox{ flat irons} = (\mbox{#orders})\cdot (\mbox{size of each order}) = n\cdot s\] the store expects to sell 110 total which can be distributed across \(x\) number of orders of a size \(s\)

We want to optimize the number of orders and order size given that we know there are two costs involved: 1) Annual Ordering Costs & 2) Annual Storage Costs

I need to wave a white flag on this problem. I honestly don’t believe there s enough information. How can we calculate the Annual Storage Cost if we have no information on how many/how long irons are kept in storage?

#6

Use integration by parts to solve the integral below. \[\int ln(9x)\cdot x^6dx\] apply integration by parts where \(u = ln(9x)\) and \(dv=x^6dx\), therefore \(v=\frac{x^7}{7}\) and \(du=\frac{1}{x}dx\) with the equation: \(\int udv = uv - \int vdu\):
\[ln(9x)\cdot \frac{x^7}{7} - \int \frac{x^7}{7}\frac{1}{x}dx \rightarrow ln(9x)\cdot \frac{x^7}{7} - \int \frac{x^6}{7}dx \rightarrow\] \[ln(9x)\cdot \frac{x^7}{7} - \frac{x^7}{49} \rightarrow ln(9x)\cdot \frac{x^7}{7} - \frac{x^7}{49} + C\]

#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. \[f(x)=\frac{1}{6x}\] What is the area under the curve from 1 to \(e^6\)? If the area = 1, the \(f(x)\) is a probability density function…. \[\int_1^{e^6} \frac{1}{6x}dx \rightarrow \frac{1}{6}ln(x)|_1^{e^6}\]

x <- 1
val1 <- 1/6*log(x)
x <- exp(6)
valexp6 <- 1/6*log(x)
A = valexp6 - val1
print( A )

## [1] 1

In conclusion, $f(x) is a probability function over this interval.