WebTest Name: (Test)

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

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

Let u = -7x => \(du = -7dx\) => \(-\frac{1}{7}du = dx\)

\(\int 4e^{-7x}dx = 4\int e^{-7x}dx = -\frac{4}{7}\int e^{u}du = -\frac{4}{7} e^{u} + C\)

  1. 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.

\(N(t) = \int \frac{dN}{dt} = \int (-\frac{3150}{t^4} - 220)dt = -\int_{t=0}^{t=1}(\frac{3150}{t^4} + 220)dt = -(-\frac{3150}{3t^3}+220t) + C = [\frac{3150}{3t^3}-220t + C]_{0}^{1} = \frac{3150}{3}-220 + C\)

for \(N(t) = 6530 = \frac{3150}{3}-220 + C\) => C = 5700

then finally, \(N(t) = \frac{3150}{3t^3} - 220t + 5700\)
  1. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x - 9\)
library(ggplot2)
f <- function(x)(2*x-9)

ggplot(mapping = aes(x = 4.5:8.5)) +
  stat_function(fun = f, size = 1, aes(color = "f(x)")) +
  labs(x = "x", y = "f(x)") +
  theme_light() +
  theme(legend.position = "bottom")

\(\int_{4.5}^{8.5}(2x-9)dx = [x^2 - 9x + C]_{4.5}^{8.5} = 16\)
  1. Find the area of the region bounded by the graphs of the given equations. \[y = x^2 - 2x - 2, y = x + 2\] Enter your answer below.

\(x^2 - 2x - 2 = x + 2 <=> x^2 - 3x - 3 = 0 <=> (x-4)(x+1) = 0 <=> x = -1,4\)

f <- function(x)(x^2 - 2*x - 2)
g <- function(x)(x+2)
i <- (as.numeric(integrate(g, lower = -1, upper = 4)[1]) - as.numeric(integrate(f, lower = -1, upper = 4)[1]))

\[\int_{-1}^{4} \bigg((x+2)-(x^2 - 2x - 2)\bigg)dx = \int_{-1}^{4} (-x^2+3x+4)dx = [\frac{-x^3}{3}+\frac{3x^2}{2}+4x]_{-1}^{4} = 20.8333333 \]

ggplot(mapping = aes(x = -5:5)) +
  stat_function(fun = g, size = 1, aes(color = "g(x)")) +
  stat_function(fun = f, size = 1, aes(color = "f(x)")) +
  labs(x = "x", y = "f(x)") +
  theme_light() +
  theme(legend.position = "bottom")
  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.

storage cost = \(3.75 * \frac{x}{2} = 1.875x\)
ordering cost = \(8.25\frac{110}{x} = \frac{907.5}{x}\)
orders = \(\frac{110}{x}\)

\(f(x) = 1.875x + \frac{907.5}{x}\) => \(f^{'}(x) = 1.875 - \frac{907.5}{x^2}\)

at \(f^{'}(x) = 0 => 1.875 - \frac{907.5}{x^2} = 0\) :

\[ \begin{cases} x = inventory \space costs = \sqrt\frac{907.5}{1.875} = 22 \\ orders = \frac{110}{x} = \frac{110}{22} = 5\end{cases} \]
  1. Use integration by parts to solve the integral below.

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

Let \(u = ln(9x)\) => \(du = \frac{1}{x}dx\) and \(dv = x^6 dx\) => \(v = \frac{x^7}{7}\)

\(\int ln(9x).x^6 dx = \int udv = uv - \int vdu = ln(9x)\frac{x^7}{7} - \int\frac{x^7}{7}\frac{1}{x}dx = ln(9x)\frac{x^7}{7} - \frac{1}{7} \int x^6dx = ln(9x)\frac{x^7}{7} - \frac{x^7}{49} + C = \frac{x^7}{7}\bigg(ln(9x) - \frac{1}{7}\bigg) + C\)
  1. Determine whether f(x) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral.

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

\(\int_{1}^{e^6} \frac{1}{6x}dx = [\frac{logx}{6}+C]_{1}^{e^6} = \frac{loge^6-log1}{6} = 1 \in [1, e^6]\)