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

∫4e^(-7x) dx

Using substitution u=−7x du/dx=d/dx(−7x)=−7 du=−7dx Hence ∫4e(−7x)dx=∫−1/7(−7)4e(−7x)dx =−1/7∫4e(−7x)(−7dx)=−1/7∫(4eu)du =−1/7∫4e(−7x)(−7dx)=−1/7∫(4eu)du =−4/7∫(eu)du=−(4/7)eu+C Replacing u with the value -7x =−(4/7)e^(−7x)+C

2.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. Contamination N of the pond will be derived by taking the integral of the change rate: dN/dt.

N=∫dN/dt=∫(−(3150/t^4)−220) dt N=−3150t−3/(−3)−220t+C=1050t−3−220t+C At t = 1 day, N = 6530 Hence substituting t = 1, 6530=1050(1)^−3−220(1)+C C=5700 Hence the function to describe the level of contamination, N at a given time t is: N(t)=1050t^−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

Area=∫4.5^8.5 (2x−9) dx Area=(8.52−9(8.5)+C)−(4.52−9(4.5)+C)=16

  1. Find the area of the region bounded by the graphs of the given equations. y=x2−2x−2,y=x+2 Plotting the 2 functions first
fun1 <- function(x)
{
  x^2 - 2*x - 2
}

fun2 <- function(x)
{
  x + 2
}

curve(fun1, -6, 6)
plot(fun2, -6, 6, add=TRUE)

Calculating the solutions, where the 2 graphs meet.

x^2−2x−2=x+2

x^2−3x−4=0

(x−4)(x+1)=0

Hence, the solutions are present at the x-values x = 4, and x = -1

Getting y for these values, the 2 points are (4,6) and (-1,1)

Now to get the area between the 2 curves, area will be the difference between the areas of the 2 curves, which means the difference in the integrals between x=-1 and x=4

$Area = _{-1}^4 (x+2) - (x^2 - 2x - 2) $ = 20.83

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 = no. of orders per year y = lot size

No. of irons produced per year = x * y, which should be greater than or equal to 110 x*y >= 110 No. of extra irons, not sold, which will be added to the cost of storage, which is $3.75 per iron. Inventory cost = 3.75 (xy - 110) + 8.75 x For minimum inventory cost, we will minimize x, that means x = 1. Hence y = 110 Hence, for minimum inventory costs, 1 order of size 110 will be optimum.

6.Use integration by parts to solve the integral below. ∫ln(9x)*(x^6)dx

As per the integration by parts formula: ∫f(x)g′(x)=f(x)g(x)−∫f′(x)g(x) Now we will take f(x) = ln(9x), hence f′(x)=1/x andwetake g′(x)=x6,henceg(x)=x7/7 ∫(x^6)ln(9x) dx = (x^7)ln(9x)/7 - ∫1/x∗(x^7)/7 dx =(x^7)ln(9x)/7 - 1/7 ∫x^6 dx =(x^7)ln(9x)/7 - x^7/49+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)

f(x)=1/(6x) ∫1(e6) 1/(6x) dx = (1/6)(ln(e^6)−ln(1)) = \((1/6)(6x1-0)\) = = 1 Hence this is a probability distribution with 100% probability under the same.