library(cubature)
  1. Use integration by substitution to solve the integral below.

\(\int{4e^{-7x}dx}\)

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

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

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN dt = ???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.

\(N(1)= 6530\)

\[\begin{split} N(t) &= \frac{1050}{t^3}-220t+C \\ N(1) &= 6530 \\ \frac{3050}{1^3}-220\times 1 +C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \end{split}\]

The level of contamination :

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

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

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

4.Find the area of the region bounded by the graphs of the given equations. y = x2 ??? 2x ??? 2, y = x + 2

Enter your answer below.

f1 = function(x){
  x+2
}
f2 = function(x){
  (x^2)- (2*x) -2
}
hcubature(f1,-1,4)$integral - hcubature(f2,-1,4)$integral
## [1] 20.83333
  1. 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.

\(x\) is the number of flat irons to be ordered.

\(Yearly\ storage\ cost = {Storage\ cost\ per\ iron} \times {Average\ number\ of\ irons\ stored} = 3.75 \times x/2 = 1.875x\)

\(Yearly\ ordering\ cost = {Cost\ of\ each\ order} \times {Number\ of\ orders} = 8.25 \times 110/x = 907.5/x\)

\(Inventory\ cost = Yearly\ storage\ cost + Yearly\ ordering\ cost = 1.875x+907.5/x = f(x)\)

To find the minimized value, differentiate and solve at \(0\):

\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \\ 1.875-\frac{907.5}{x^2} &= 0 \\ 1.875&= \frac{907.5}{x^2} \\ 1.875x^2&= 907.5 \\ x^2&= \frac{907.5}{1.875} \\ x&= \sqrt{\frac{907.5}{1.875}} \\ x&=\sqrt{484} \\ x&=22 \end{split} \] Each order should contain \(22\) flat irons as lot size, therefore, it should be \(110/22=5\) orders.

  1. Use integration by parts to solve the integral below.

\(\int{ln(9x) \times x^6 dx}\) Let \(u= ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\).

Let \(\frac{dv}{dx}=x^6\), then \(v = \int{x^6 dx} = \frac{1}{7}x^7\).

Using the formula for integration by parts: \(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)

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

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.

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

\[ \begin{aligned} \int^{e^6}_{1} 1/6x \,dx = \big[\ln(6x) \big]^{e^6}_1 \\ = \ln(6e^6) - \ln(6) \\ \ln(6) + \ln(e^6) - \ln(6) = 6 \neq 1 \end{aligned} \]

It is not a probability distribution because the area is not 1.