Question 1

Use integration by substitution to solve the integral below. \[\int4e^{-7x}dx\] \[u = -7x\] \[du = -7dx\] \[dx = \frac{du}{-7}\] \[4\int e^u \frac{du}{-7}\] \[-\frac{4}{7} * e^{-7x} + C\]

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

\[{N}(t) = -\frac{3150t^-3}{-3} - 220t = 6530\] \[{N}(t) = \frac{1050}{t^3} - 220t = 6530\] \[{N}(1) = \frac{1050}{1^3} - 220*1 = 6530\] \[{N}(1) = 6530 - 1050 + 220 = 5700\] \[{N}(t) = \frac{1050}{t^3} - 220t + 5700\]

Question 3

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

\[\int_{4.5}^{8.5}2x-9dx = x^2 - 9x \Biggr|_{4.5}^{8.5}\]

upval <- 8.5
lowval <- 4.5
(upval^2 - 9*upval) - (lowval^2 - 9*lowval)
## [1] 16

Question 4

Find the area of the region bounded by the graphs of the given equations. \[y = x^2 - 2x -2,\space y = x + 2\] \[y = x^2 - 2x -2 = x + 2\] \[y = x^2 - 3x -4\] \[(x-4)=0, \space (x+1)=0\] \[x=4, \space x=-1\]

\[\int_{-1}^{4}(x + 2 dx = \frac{x^2}{2} + 2x) - (x^2 - 2x -2 dx = \frac{x^3}{3} - \frac{2x^2}{2} - 2x) \Biggr|_{-1}^{4}\]

upval <- 4
lowval <- -1

# fx = x + 2
((((upval^2)/2) + (2*upval)) - (((lowval^2)/2) + (2*lowval))) -
# fx = x^2 -2x -2
  ((((upval^3)/3) - ((2*(upval^2))/2) - (2*upval)) - 
  (((lowval^3)/3) - ((2*(lowval^2))/2) - (2*lowval)))
## [1] 20.83333
curve(x^2 -2*x-2, -1, 4)
curve(x+2, add = TRUE, -1,4)

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.

Let \(x\) number of irons in a lot size, with half of orders with storage cost

\(3.75 * \frac{x}{2}\) \(8.25 * \frac{110}{x}\)

\(\int1.875x + \frac{907.5}{x}\)

\(1.875 + \frac{907.5}{x^2} = 0\)

\(x = \sqrt{\frac{907.5}{1.875}}\)

Lot size: \(x = 22\)

Number of Orders: \(\frac{110}{x} = \frac{110}{22} = 5\)

Question 6

Use integration by parts to solve the integral below. \[\int ln(9x) * x^6dx \] \(u = ln(9x) , v = x^6\)

\(u' = \frac{1}{x}\)

\(\int v dx = \frac{x^7}{7}\)

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

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

\(ln(9x) * \frac{x^7}{7} - \int \frac{x^6}{7} dx\)

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

\(ln(9x) * \frac{x^7}{7} - \frac{x^7}{49} + C\)

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. \[f(x)=\frac{1}{6x}\] \[g(x)=\int_{1}^{e^6}\frac{1}{6x} dx = \frac{1}{6}ln(x)\Biggr|_{1}^{e^6}\] \[\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1)\] It’s a probability density function

(1/6) * log(exp(6)) - ((1/6) * log(1))
## [1] 1