Question 1:

Use integration by substitution to solve the integral below

\[\int4e^{-7x}dx\\ ............... \\ u = -7x \\ du = -7dx \\ dx = \frac{du}{-7}\\ \int4e^{u}\frac{du}{-7} \\ -\frac{4}{7}\int e^{u}du = \\ -\frac{4}{7}e^{u} + C \\ -\frac{4}{7}e^{-7x} + C \]

Using integration by substitution for \(\int4e^{-7x}dx\) we get \(\frac{4}{7}e^{-7x} + C\)

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.

To find the function \(N(t)\) we can integrate \(\frac{dN}{dt}\)

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

First, integrate each term separately

First term
\(\int\frac{3150}{t^4} dt\) \(3150\int{t^4}dt\) \(3150 (\frac{t^-3}{-3}) + C\) \(-1050 * t^{-3}+ C\)

Second term \(\int-220dt\) \(-220 \int dt\) \(-220t + C\)

Combining the terms to get N(t) \(N(t) = -1050t^{-3} - 220t + C\)

Using the information given in the problem we can solve for C to complete the function \(6530 = -1050(1)^{-3} - 220(1) + C\) \(6530 = -1270 + C\) \(C = 7,800\)

A function to estimate the level of contamination is \(N(t) = \frac{-1050}{t^{3}} - 220t + 7800\)

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.

Adding the area of each rectangle we get a total area of 16.

Question 4:

Find the area of the region bounded by the graphs of the given equations. \(y=x^2-2x-2\) and \(y=x+2\)

First we can find the points were the two functions cross by plotting the lines or solving for the values of x

\[x^2-2x-2 = x+2 \\ x^2 - 3x - 4 = 0\]

Using the quadratic equation we get x = -1 or x = 4

The area bound by the two graphs is

\(\int\limits_{-1}^4((x + 2) - (x^2 - 2x - 2))dx\) \

\(\int\limits_{-1}^4(-x^2 + 3x + 4)dx\)

\(\left [\begin{array} --\frac{x^3}{3} + \frac{3x^2}{2} + 4x \end{array} \right ]_{-1}^4\)

\(\left [\begin{array} -\frac{4^3}{3} + \frac{3(4)^2}{2} + 4(4) \end{array} \right ] - \left [\begin{array} -\frac{(-1)^3}{3} + \frac{3(-1)^2}{2} + 4(-1) \end{array} \right ]\)

\(\left [\begin{array} -\frac{64}{3} + \frac{48}{2} + 16 \end{array} \right ] - \left [\begin{array} -\frac{-1}{3} + \frac{3}{2} + -4 \end{array} \right ]\)

\(\left [\begin{array} -\frac{128}{6} + \frac{144}{6} + \frac{96}{6} \end{array} \right ] - \left [\begin{array} -\frac{2}{6} + \frac{9}{6} + -\frac{24}{6} \end{array} \right ]\)

\(\frac{112}{6} - -\frac{13}{6} = \frac{125}{6}\)

\(= 20.833\)

f <- function(x) {
  (-1*x)**2 - (3*x) - 4
}
upper <- 4
lower <- -1

integrate(f, -1, 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.

*(I wasn’t totally sure how to answer this question but in doing some research I came upon the economic order quantity formula that a company can use to minimze inventory cost)

\(Q = \sqrt\frac{2DS}{H}\)

D <- 110
S <- 8.25
H <- 3.75

Q <- sqrt((2*D*S)/H)
Q
## [1] 22

The lot size or order quantity is 22. The number of orders per year can be found by taking the demand / the lot size \(\frac{Q}{D}\). In the case above this would be 5 orders per year to minimize inventory costs

Quesiton 6:

Use integration by parts to solve the integral below

\[\int ln(9x) * x^6dx \\ \int f(x)g'(x) = f(x)g(x) - \int f'(x)g(x) \\ f'(x) = \frac{1}{x} , ~g(x) = \frac{x^7}{7} \\ \frac{x^7 ln(9x)}{7} - \int \frac{x^6}{7} dx \\ \int \frac{x^6}{7} dx = \frac {1}{7} \int x^6 dx = \frac{1}{7} * \frac{x^7}{7} = \frac{x^7}{49} \\ \int ln(9x) * x^6dx = \\ \frac{x^7 ln(9x)}{7} - \frac{x^7}{49} + C\\ \]

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.

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

To determine if f(x) is a probability density function we check two conditions

  1. \(f(x)≥0\) for all x in the given interval.

    • Since all values of are positive in the interval \([1, e^6]\) f(x) will be \(\geq 0\)

  2. The total area under the curve of f(x) over the interval \([1, e^6]\) is equal to 1.

    • The area under the curve can be found by calculating the integral of the function. As shown below the area under the curve is 1.

\[\int\limits_{1}^{e^6} \frac{1}{6x}dx \\ = \frac{1}{6} \int\limits_{1}^{e^6} \frac{1}{x}dx \\ \frac{1}{6}(ln(e^6) - ln(1)) \\ \frac{1}{6}(6-0) \\ = 1\]

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

upper <- exp(1)**6
lower <- 1

integrate(g,lower, upper)
## 1 with absolute error < 9.3e-05

Both conditions are met so \(f(x) = \frac{1}{6x}\) is a probability density function