1. Use integration by substitution to solve the integral below.
knitr::include_graphics("https://raw.githubusercontent.com/maharjansudhan/DATA605/master/IMG_1620.jpg")

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.

Answer:

Here

???(???3150/t^4)???220dt = -3150 * (t???3/-3) - 220t + C = x

After 1 day, x = 6530, solving for C

C <- 6530 + 3150 * ((1^-3)/-3) + 220*(1)
C
## [1] 5700

N(t) = -3150 * (t^???3/-3) - 220t + 5700 = 1050 t^???3 - 220t + 5700

  1. Find the total area of the red rectangles in the figure below, where the 3. equation of the line is f ( x ) = 2x-9.

Answer: Here, f(x) = 2*x - 9

knitr::include_graphics("https://raw.githubusercontent.com/maharjansudhan/DATA605/master/download.jpg")

# According to the graph

Area = (5.5 - 4.5) * (1 - 0) + (6.5 - 5.5) * (3 - 0) + (7.5 - 6.5) * (5 - 0) +(7.5 - 6.5) * (7 - 0)
Area
## [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

Answer:

# to calculate the area
X1 = c(-10:10)
Y1 = X1^2 - 2*X1 - 2
X2 = c(-10:10)
Y2 = X2 + 2
plot(X1,Y1,type="l", xlab="X1,X2", ylab="Y1, Y2")
lines(X2,Y2,type="l")

With these we can solve for x and y

x + 2 = x^2 - 2x - 2 0 = x^2 - 3x - 4 0 = (x -4) (x + 1)

x = 5, -1

for y, - y = x + 2

y = 7, 1

Therefore, the points are (5,7) and (-1,1).

  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.

Answer: Here,

Suppose, x be number of orders peryear y be lot size per order

Then x * y = total number of irons ordered for the year x * y >= 110, there should be some sales so x should be greater that or equal to 1

fixed cost = $8.25 store cost = $3.75 units sells expected = 110

Now, cost of inventory per year = 3.75 (xy - 110) + 8.75x

# calculate minimum cost when selling unit is 1 
min_cost <- 3.75 * ((110)*(1) - 110) + 8.75*(1)
min_cost
## [1] 8.75
  1. Use integration by parts to solve the integral below.
knitr::include_graphics("https://raw.githubusercontent.com/maharjansudhan/DATA605/master/IMG_1621.jpg")

7.Determine whether f ( x ) is a probability density function on the interval 1, e6 . If not, determine the value of the definite integral.

f(x) = 1/6x

Answer:

Here,

1/6 * ln x from [1, e^6], =

# to calculate probability density function 
probability_density_fn <- (1/6) * log(exp(6)) - ((1/6) * log(1))
probability_density_fn
## [1] 1