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

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

Let u = -7x

\[ \dfrac{du}{dx} = -7 \]

\[ du = -7dx \]

\[ dx = \dfrac{du}{-7} = -\dfrac{1}{7}du\]

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

\[ \dfrac{4}{7}e^{u} = \dfrac{4}{7}e^{-7x} + C\]

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \[\dfrac{dN}{dt} = -\dfrac{3150}{t^4} - 220\] 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.

\[\dfrac{dN}{dt} = -\dfrac{3150}{t^4} - 220\]

\[dN = -\dfrac{3150}{t^4} - 220 dt\]

\[dN = \int-\dfrac{3150}{t^4}dt - \int220 dt\]

\[dN = \int-3150t^{-4}dt - \int220 dt\]

\[N(t) = \dfrac{3150}{3 \; t^3} - 220t \: + \: C\]

At t = 1, N(1) = 6350

\[N(1) = \dfrac{3150}{3} - 220 \: + \: C = 6350\]

Therefore,

\[1050 - 220 = 830 = C\]

\[N(t) = \dfrac{3150}{3 \; t^3} - 220t \: + \: 830\]

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

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

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

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

((8.5)*(8.5)-(9*8.5))-((4.5)*(4.5)-(9*4.5))
## [1] 16
  1. Find the area of the region bounded by the graphs of the given equations.

y=x2-2x-2

y=x+2

eqn1 <- function(x){
  return(x**2 -2*x - 2)
}

eqn2 <- function(x){
  return(x + 2)
}

x <- seq(-2, 5, 1)
y <- eqn2(x)

plot(eqn1, from = -2, to = 5, line = "l", col = 'blue')
## Warning in plot.window(...): "line" is not a graphical parameter
## Warning in plot.xy(xy, type, ...): "line" is not a graphical parameter
## Warning in axis(side = side, at = at, labels = labels, ...): NAs introduced
## by coercion

## Warning in axis(side = side, at = at, labels = labels, ...): NAs introduced
## by coercion
## Warning in box(...): "line" is not a graphical parameter
## Warning in title(...): NAs introduced by coercion
lines(x,y, col = 'green')

Enter your answer below.

\[x^2 - 2x -2 = x+2\]

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

\[x = 1, x=-4\]

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

\[ (\dfrac{x^2}{2})|_{-1}^{4} - (\dfrac{x^3}{3}-x^2-2x)|_{-1}^{4} \]

(8-.5)-((((4*4*4)/3) - 16-8) - (-.3333 -1 + 2))
## [1] 10.83337
  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.

\[Cost = 8.25 \: * \: order \:+ \: 3.75 \: * \: order \: * \: lot \]

\[C = 8.25 \: o \:+ \: 3.75 \:eo\]

\[ \dfrac{de}{do} 8.25 + 3.75eo\: = 8.25 + 3.25e\]

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

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

Let u = 9x

du = dx

v =

dv = x^6lx

\[ \int ln(u) * (v) = uv - \int v \:du \]

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

\[\dfrac{x^7}{7} ln(9x) - \dfrac{1}{7} \]

  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.

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

To tell if a function is a probability density function, the integral of the pdf must be equal to 1.

\[\int_{1}^{e^6} \dfrac{1}{6x}\]

\[-\dfrac{1}{6}\int_{1}^{e^6} \dfrac{1}{x}\]

\[-\dfrac{1}{6}ln(x)|_{1}^{e^6}\]

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