1. Use integration by substitution to solve the integral below. \[∫4 e−7 dx\] First, differentiate with respect to x: \[ du/dx = -7\] \[du=-7dx\] \[dx=-1/7 du\]

Let: \[ u = -7x\]

Substitute u and dx into the integral: \[∫4e^u(-1/7)(du) = (-4/7)∫e^u(du)\] Then integrate and substitute for u: \[(-4/7)(e^u)+C\] \[(-4/7)(e−7x)+C\]

  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.

We can establish a function for the level of conatmination using the rate of contamination which we were given. First, we have to integrate the contamination rate like so: \[dN/dt = - 3150/t^4 -220\] \[∫ dN/dt(dt) = ∫(−3150/t^4 -220)dt\] \[N(t)=∫(−3150/t^4 -220)\] Take the integral of that function: \[N(t) = 3150/3t^3 -220t + C\] And then substitute with the day one level of contamination, 6530, to solve for C

\[ N(1) = 3150/3(1)^3 -220(1) + C = 6530\] \[C = 6530+220-(3150/3) = 5700\]

The function to estimate the level of contamination: \[N(t) = 3150/3t^3 -220t + 5700\]

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

First rectangle: - height: 1 - base: 1 - area = 1 x 1 = 1

Second: - height: 3 - base: 1 - area = 3 x 1 = 3

Third: - height: 5 - base: 1 - area = 5 x 1 = 5

Fourth: - height: 7 - base: 1 - area = 7 x 1 = 7

Combined area = 1 + 3 + 5 + 7 = 16

  1. Find the area of the region bounded by the graphs of the given equations. \[ y = x^2 - 2x - 2, y = x + 2\]
curve(x^2 - 2*x - 2, -2, 5, col = "red")
curve(x + 2, -2, 5, add=T, col="blue")

After graphing the two functions, we can see that we are trying to determine the area between two intersections at or near x=-1 and x=4.

We need to set two equtions equal to one another like so: \[x^2 - 2x - 2 = x + 2\] Then simplify: \[x^2 - 3x -4 = 0\] We can factor this out like so: \[(x-4)(x+1)=0\]

And solve for x: \[x=4\] \[x=-1\] Now that we have found the x values for each of the two intersection points, we can plug them back into the equation to solve for y:

Solving for y using x=4: \[(4)^2-2(4)-2=16-8-2=6\] Coordinate is: (4,6)

Solving for y using x=-1: \[(-1)^2-2(-1)-2=1+2-2=1\] Coordinate is: (1,1)

Now, we know that the two coordinates are (4,6) and (1,1).

To find the area between these two points, we can find the integral of the difference of the functions over the interval [-1,4]:

Area: \[∫((x+2)-(x^2-2x-2))dx\]

Which simplifies to the following: \[Area(from[-1,4])=-(x^3/3)+(3x^2)/2+4x\] We can substitute and subtract the two equations: \[Area(from[-1,4])=[-(((4)^3/3)+(3(4)^2)/2+4(4)] -[-(((1)^3/3)+(3(1)^2)/2+4(1)] \] \[ Area = 365/6\] The area is equal to about 60.83.

  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.

We are given the following:

Total annual demand for flat irons = 110 Cost to place an order = $8.25x (x is added because this is for each order) Cost to store one flat iron for one year = $3.75 Annual storage = 3.75

Let: x = # of orders per year n = lot size (number of flat irons per order)

The holding cost of the irons would be as follows: Holding cost = \[3.75(110/x)\]

This is because each order, n, costs $3.75 per flat iron and there are 110/x orders per year.

Meaning that the total cost is: Cost = \[ 8.25x + 3.75(110/x)\]

We can take the derivative to minimize costs: \[dC/dx=8.25 - 3.75(110/x^2)\] Then set it equal to 0 and solve, getting us x=5:

\[0=8.25 - 3.75(110/x^2)\] \[x=5\]

We can use that to find the lot size like so: \[110/(5)=22\] It would take 5 orders to minimize inventory costs, and a lot size of 22, while still meeting the demands for 110 flat irons yearly.

  1. Use integration by parts to solve the integral below: \[ ∫ln(9x)x^6dx\]

\[∫fg'=fg-∫lf'g\] \[f=ln(9x)\] \[f'=1/x\] \[g'=x^6\] \[g=x^7/7\] Then substitute like so:

\[∫ln(9x)x^6dx=ln(9x)((x^7)/7)-∫(1/x)((x^7)/7)\] \[(ln(9x)x^7)/7-(x^7/49)\] \[(7x^7ln(9x)-x^7)/49\] \[x^7(7ln(9x)-1)/49 + C\] 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. \[f(x)=1/6x\]

We need to make sure that over the interval, the area under the function’s curve is equal to 1. To do this, we need to integrate the function over its domain and check if it equals 1.

\[∫f(x)dx=∫(1/6x)dx\] \[∫(1/6x)dx=(1/6)∫(1/x)dx\]

\[(1/6)ln(x)=(1/6)[ln(e^6)-ln(1)] = (1/6)[6-0] = 1\]

Because the integral over the function’s domain equals 1, the function is a valid probability density function.