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

∫4e^−7xdx

Take the constant out: ∫a · f(x) dx = a · ∫f(x) dx

= 4 · ∫ e ^-7x dx

Apply u- substitution

= 4 · ∫ − 1/7e^u du

Take the constant out: ∫a · f(x) dx = a · ∫f(x) dx

= 4(-1/7 · ∫ e^u du)

Use the common integral: ∫ e^u du = e^u

= 4(-1/7e^u)

Substitute back u = −7x

= 4(-1/7e^-7x)

Simplify 4(-1/7e-7x):-4/7e-7x

= -4/7e^-7x

Add a constant to the solution

= -4/7e^-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.

dN/dt = 3150/t^4 - 220

dN/dt=−3150/t^-4−220

N(t)=∫−3150/t^-4−220dt

=−3150t^−3/−3−220t+c

N(t)=1050/t^3−220t+c

Given level of contamination after 1 day was 6530 bacteria per cubic centimeter:

6530=1050/1^3−220(1)+c

6530=830+c

c=5700

Thus:

N(t)=1050/t^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.

Based on the above figure, the rectangles start at x = 4.5 and end at x = 8.5 We will integrate the function at these points to find the area.

∫^8.5 4.5 * 2x−9dx

= x2−9x|8.5 4.5

Area=(8.52−9(8.5))−(4.52−9(4.5))

Area=(72.25−76.5)−(20.25−40.5)

Area=(−4.25)−(−20.25)

Area of the red retangle is 16

Also: Each square in the graph has an area of 1. Each rectangle has a width of 1. Counting each rectangle left to right the areas are :

(Area=1+3+5+7)
## [1] 16
  1. Find the area of the region bounded by the graphs of the given equations. y = x^2 - 2x - 2, y = x + 2
#Find area in-build function
f1 = function(x) {x + 2}
f2 = function(x) {x^2 -2*x -2}

#Find the difference between areas under the curve
area1 <- integrate(f1, -1, 4)
area2 <- integrate(f2, -1, 4)
(area <- round((area1$value - area2$value),4))
## [1] 20.8333
# plot
curve(x^2 - 2*x - 2, -5, 5, col = "green")
curve(x + 2, -5, 5, add=T, col="orange")

  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.

Number of orders (x) * Lot Size (n) = 110 C = costs

C = 8.25x + 3.75 ∗ 110/x/2

C = 8.25x + 3.75 ∗ 55/x

C = 8.25x + 206.25/x

In order to minimize costs, set derivative = zero

C′ = 8.25 − 206.25/x^2

0 = 8.25 − 206.25/x^2

206.25/x2 = 8.25

206.25 = 8.25x^2

206.25/8.25 = x^2

25 = x^2

Thus, the number of orders per year that will minimize inventory costs is 5. The lot size is 22 because 22 x 5 = 110

  1. Use integration by parts to solve the integral below

∫ln(9x) * x^6 * dx

∫fg′ = fg − ∫f′g

f = ln(9x)

f′ = 1/x

g′ = x^6

g = x^7/7

Substitute into formula

∫ln(9x) ∗ x^6 = ln(9x) ∗ x^7/7 − ∫1/x ∗ x^7/7

= ln(9x) ∗ x^7/7 − ∫x^6/7

= ln(9x) ∗ x^7/7 − x^7/49

= 7 ∗ ln(9x) ∗ x^7/49 − x^7/49

= (7x^7 ∗ ln(9x))−x^7/49

= x^7(7ln(9x)−1)/49 + c

  1. Determine whether f ( x ) is a probability density function on the interval [1,e^6]. If not, determine the value of thedefinite integral. f ( x ) = 1/6x

∫e611/(6x)dx 1/6 * ln x from [1, e6], =

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

f(x) is a probability density function on the interval [1, e6] - with the area of the curve as 1 (100% probabily for the interval)