Ques-1: Use integration by substitution to solve the integral below:

\(\int\) \(4e^{-7x}\) \(dx\)

Ans-1:

Let, \(u = -7x\)

By taking derivative of u with respect to x we get,

\(du = -7dx\)

or, \(dx = -du/7\)

Now, we have, \(\int\) \(4e^{-7x}\) \(dx\)

Putting \(-7x = u\) and \(dx = -du/7\) we get,

\(-4/7\) \(\int\) \(e^{u}\) \(du\)

\(=\) \(-4/7\) \(e^{u}\) + \(C\)

\(=\) \(-4/7\) \(e^{-7x}\) + \(C\)

Ques-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

Ans-2:

Given rate, dN/dt = -3150/ \(t^{4}\) - 220, integrating the rate equation with respect to time we get,

\(\int\) \(dN\) = -3150 \(\int\) \(t^{-4}\) dt - 220 \(\int\) dt

\(N(t)\) = \(3150/3\) \(t^{-3}\) - \(220t + C\)

\(N(1)\)=\(6530\)= \(3150/3\) \(1^{-3}\) - \(220 + C\) [for, t=1, N(1)=6530]

\(C= 6530-1050+220\)

\(C = 5700\)

Therefore, the function will be: \(N(t)\) = \(1050\) \(t^{-3}\) - \(220t + 5700\)

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

Ans-3:

From the given plot we get the lower limit= 4.5 and the upper limit =8.5. Now, by integrating the given line function within this limits we get the area. Below is the R code for finding the area by doing integration.

# define the area function
area_func<-function(x){
  
  return(2*x-9)
}

# Integrate the function over the limits
result<-integrate(area_func, lower = 4.5, upper = 8.5)

result$value
## [1] 16

Ques-4: Find the area of the region bounded by the graphs of the given equations.

\(y= x^{2}−2x−2\) , \(y=x+2\)

Ans-4:

Solve the given two equations for x values:

\(x+2 = x^{2}−2x−2\)

or, \(x^{2}-3x-4 = 0\)

or, \(x^{2}-4x+x-4 = 0\)

or, \(x(x-4)+1(x-4)=0\)

or, \((x+1) (x-4)=o\)

Therefore, \(x= -1\) and \(x=4\)

define two functions

func_1 <- function(x){
  return(x^2 - 2*x -2)
}

func_2 <- function(x){
  return(x +2)
}

plot graphs

plot(func_1, -10, 10,ylab = "y")
plot(func_2, -10, 10, add = TRUE)

From the plot above, it is seen that the x values at the two curves intersection points are -1 and 4 respectively. We are going to use this values as the lower and upper limits of the integration to find the desired area.

Find area of the intersection

# define function for intersection
inter_func <- function(x){
    return((x+2) - (x^2 - 2*x -2))
}

# find area for bounded by found x limits
area <- integrate(inter_func, lower = -1, upper = 4)
area$value
## [1] 20.83333

Ques-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.

Ans-5:

Let’s consider, order quantity, q annual demand, d = 110 (targeted sale number) holding cost per unit per year, h= 3.75 and fixed cost per order, k =8.25

# Define the function for calculating total annual inventory cost
total_inventory_cost <- function(q, d= 110, h = 3.75, k= 8.25) {
 
# Calculate annual ordering cost
annual_order_cost<- k* d / q
  
# Calculate annual holding cost
annual_holding_cost<-h * q / 2 # assuming that the average inventory level is half of the order quantity.
  
# Calculate total annual inventory cost
  total_annual_inventory_cost= annual_order_cost +  annual_holding_cost
  
# Return total cost
 return(total_annual_inventory_cost)
}

# Optimization techniques to find the optimal order quantity
optimized_result <- optimize(total_inventory_cost, interval = c(1, 110), d = 110, h = 3.75, k = 8.25)

# Minimum lot size 
optimized_result$minimum
## [1] 22.00002
# Minimum total inventory cost
optimized_result$objective
## [1] 82.5

Therefore, the lot size that will minimize inventory costs is 22 flat irons, and the number of orders per year will be 110 /22 = 5 orders per year.

Ques-6: Use integration by parts to solve the integral below:

\(\int\) \(ln(9x)x^{6}dx\)

Ans-6:

Let, \(u = ln(9x)\) and, \(v = x^{6}\)

So, \(du = 1/x dx\)

and \(dv = 1/7x^{7}dx\)

Now, \(\int\) \((uv)dx\) = \(ln(9x)∗1/7x^{7}\)− \(\int\)\(1/x* 1/7* x^{7}dx\)

\(\int\) \((uv)dx\) = \(ln(9x)∗1/7x^{7}\)−1/7\(\int\)\(x^{6}dx\)

\(\int\) \((uv)dx\) = \(ln(9x)∗1/7x^{7}\)−\(1/49x^{7}+C\)

Ques-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.

Ans-7:

\(f(x)= 1/6x\)

\(\int_{1}^{e^{6}}\) \(1/6xdx\)

\(=\) \(1/6\)∗\(lnx\)\(|_{1}^{e^{6}}\)

\(=1/6ln(e^{6})−1/6ln(1)\)

\(=1/6[6−0]=1\)

The function, \(f(x) = 1/(6x)\) is non-negative for all \(x\) in \([1, e^{6}]\) and the integral of the function over the given limit is equal to 1. Therefore, the function, f(x) is a probability density function on the given interval.