DATA 605 HW 13 YINA QIAO

Problem 1:

Use integration by substitution to solve the integral below.

\[\int 4e^{-7x} \, dx\] \[ \begin{split} \int{4e^{-7x}dx} \\ u = -7x \\ du = -7dx \Rightarrow dx = \frac{du}{-7} \end{split} \] by substitution \[ \begin{split} \int{4e^u \frac{du}{-7}} \\ -\frac{4}{7}\int{e^udu}\\ -\frac{4}{7}e^u + C\\ -\frac{4}{7}e^{-7x} + C\\ \end{split} \]

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. \[\begin{split} \frac{dN}{dt} = N'(t) = \frac{3150}{t^4}-220 \\ N(t) = \int{\frac{3150}{t^4}-220dt} \\ = \frac{1050}{t^3}-220t + C \end{split}\]

$$ We are given \(N(1) = 6530\) so we can plug that in to solve for \(C\)

\[ \begin{split} 6530 = \frac{1050}{1^3}-220 * 1 + C \\ 6530 = 830 + C \\ C = 5700 \end{split} \] This gives us

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

Problem 3:

Find the total area under the curve defined by \(f(x) = 2x - 9\).

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.

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

funOne <- function(x){x^2 - 2*x - 2}
funTwo <- function(x){x+2}
curve(funOne, -2,5)
curve(funTwo,-2,5,add=T)

To find the area we take the integral of the upper function minus the lower function

\[\begin{split} \int_{-1}^{4}{(x^2 - 2x -2) - (x+2)dx} \\ \int_{-1}^{4}{x^2 - 3x -4dx} \end{split}\]

funDiff <- function(x){x^2 - 3*x - 4}
integrate(funDiff, -1,4)

## -20.83333 with absolute error < 2.3e-13

The area bound by the graphs of the given equations is 20.83

Problem 5:

A beauty supply store expects to sell 110 flat irons during the next year. Find the lot size and the number of orders per year that will minimize inventory costs.

D <- 110  # Annual demand
S <- 8.25  # Fixed cost per order
H <- 3.75  # Holding cost per unit per year
Q_star <- sqrt((2 * D * S) / H)
N_star <- D / Q_star
list(Q_star = Q_star, N_star = N_star)

## $Q_star
## [1] 22
## 
## $N_star
## [1] 5

The lot size should be 22 and numbers of orders per year is 5.

Problem 6:

Solve the integral using integration by parts.

\[\int \ln(9x) \cdot x^6 \, dx \] \[\begin{split} \int{udv} = uv - \int{vdu} \\ u = ln(9x),du = \frac{1}{x},v = \frac{1}{7}x^7,dv = x^6\\ ln(9x) \frac{1}{7}x^7 - \int{\frac{1}{7}x^7\frac{1}{x}dx} \\ ln(9x) \frac{1}{7}x^7 - \frac{1}{7}\int{x^6dx} \\ ln(9x) \frac{1}{7}x^7 - \frac{1}{7} \times \frac{1}{7}x^7 + C \\ ln(9x) \frac{1}{7}x^7 - \frac{x^7}{49} + C \end{split}\]

Problem 7:

Determine whether \(f(x) = \frac{1}{6x}\) is a probability density function on the interval \([1, e^6]\).

f <- function(x) 1 / (6 * x)
integrate(f, 1, exp(6))

## 1 with absolute error < 9.3e-05