Question 1:

Use integration by substitution to solve the integral below:

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

Steps:

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

\[ \int 4 e^{u} \ \frac{du}{-7} \]

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

\[ -\frac{4}{7}e^{u} + C \]

\[ -\frac{4}{7}e^{-7x} + C \]

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.

\[ \int -\frac{3150}{t^4} - 220 dt \]

\[ \int -3150t^{-4} - 220 dt \]

\[ \frac {3150}{3} t^{-3} - 220t + C \]

\[ \frac{1050}{t^3} - 220t + C \]

Solve for C:

\[ 6530 = \frac{1050}{1^3} - 220(1) + C \]

\[ 6530 = 1050 - 220 + C \]

\[ C = 5700 \]

Function:

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

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.

library(grid)
library(png)

img <- readPNG("~/photoHW13.png")

plot(1, type = "n", axes = FALSE, xlab = "", ylab = "", xlim = c(0, 8), ylim = c(0, 10))

img_grob <- rasterGrob(img, width = unit(1, "npc"), height = unit(1, "npc"))
grid.draw(img_grob)

\[ \int_{4.5}^{8.5} 2x - 9 \ dx \]

\[ [(8.5^2) - (9 * 8.5)] \ - \ [(4.5^2) - (9 * 4.5)] = [72.25 - 76.5] \ - \ [20.25 - 40.5 = -4.25 \ - \ -20.25 \ = \ 16 \]

Answer: 16 units squared

Another way to determine this value is to count the number of squares in the picture. At x-value 5, you have one square. At x-value 6, you have three squares. At x-value 7, you have five squares. At x-value 8, you have seven squares. One + Three + Five + Seven = Sixteen.

Another way to determine this value is to calculate the area of a triangle that goes from x = 4.5 to x = 8.5, so a width of 4, and a height from y = 0 to y = 7, so a height of 7. \(A_{triangle} = \frac{1}{2} b h\). A = (1/2)(4)(7) = (1/2)(28) = 16.

Question 4:

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

\(y = x^2 - 2x - 2, \ y = x + 2\)

  1. Find the intersection points:

    \(x^2 - 2x - 2 = x + 2\)

    \(x^2 - 3x - 4 = 0\)

    \((x - 4)(x + 1) = 0\)

    Intersections at x = 4 and x = -1.

  2. Integrate from x = -1 to x = 4.

\[ \int_{-1}^{4} x + 2 \ dx \ - \int_{-1}^{4} x^2 - 2x - 2 \ dx \]

\[ [ \ [\frac{1}{2}(4)^2 + 2(4)] - [\frac{1}{2}(-1)^2 + 2(-1)] \ ] \ - \ [ \ [\frac{1}{3}(4)^3 - (4)^2 - 2(4)] - [\frac{1}{3}(-1)^3 - (-1)^2 - 2(-1)] \ ] \]

\[ = [ \ [8 + 8] - [\frac{1}{2} - 2] \ ] \ - \ [ \ [\frac{64}{3} - 16 - 8] - [-\frac{1}{3} - 1 + 2] \ ] \]

\[ = [ \ [16] - [-\frac{3}{2}] \ ] \ - \ [ \ [-\frac{8}{3}] - [\frac{2}{3}] \ ] \]

\[ = [\frac{35}{2}] \ - \ [-\frac{10}{3}] = [\frac{35}{2}] \ + \ [\frac{10}{3}] = \frac{125}{6} \]

Answer: 125/6 = 20.833333 units squared

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.

Number of orders * lot size = 110

You will make a certain number of orders (x) at a certain quantity (lot size) to have a total of 110 flat irons, which you will then sell. Your orders are your business orders for inventory. Lot size = 110 / x.

\(y = (3.75 \ * \ \frac{110/x}{2}) + (8.25 \ * \ x)\)

\(y = \frac{206.25}{x} \ + \ 8.25x\)

To minimize costs, find the value of x when the derivative equals 0.

\(y' = \frac{-206.25}{x^2} + 8.25\)

\(0 = \frac{-206.25}{x^2} + 8.25\)

\(8.25 = \frac{206.25}{x^2}\)

\(x^2 = \frac{206.25}{8.25}\)

\(x = \sqrt {\frac{206.25}{8.25}}\)

\(x = 5\)

Answer: 22 lot size and 5 orders

Question 6:

Use integration by parts to solve the integral below:

\[ \int ln(9x) \ * \ x^6 \ dx \]

Integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

For this problem:

\[ u = ln(9x), \ du = \frac{1}{x}dx, \ v = \frac{1}{7}x^7, \ dv = x^6 dx \]

\[ ln(9x) * \frac{1}{7}x^7 - \int \frac{1}{7}x^7 \frac{1}{x} dx \]

\[ \frac{1}{7} ln(9x)x^7 - \frac{1}{7} \int x^6 dx \]

\[ \frac{1}{7} ln(9x) x^7 - \frac{1}{7} (\frac {1}{7} x^7) + C \]

\[ \frac{x^7}{7} (ln(9x) - \frac{1}{7}) + C \]

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.

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

If f(x) is a probability density function, then the area under the curve must equal 1. We will find the area under the curve from x = 1 to x = \(e^6\) first.

\[ \int_{1}^{e^6} \frac {1}{6x} dx \ = \ \int_{1}^{e^6} \frac{1}{6}x^{-1} dx \]

\[ = \frac {1}{6} ln(e^6) - \frac{1}{6}ln(1) = \frac {ln(e^6) - ln(1)}{6} = (6 - 0) / 6 = 1 \]

Since the area under the curve is equal to 1, this is a probability density function.