1. Use integration by substitution to solve the integral below - \(\int { 4{ e }^{ -7x }dx }\)

Ans: Let u = -7x

du = -7dx

=> dx = -du/7

\(\int { 4{ e }^{ -7x }dx } \quad =\quad \frac { -4 }{ 7 } \int { { e }^{ u }du } \\ \quad \quad \quad \quad \quad =\quad \frac { -4 }{ 7 } { e }^{ u }+C\\ \quad \quad \quad \quad \quad =\quad \frac { -4 }{ 7 } { e }^{ -7x }+C\)

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of - \(\frac { dN }{ dt } =-\frac { 3150 }{ { t }^{ 4 } } -220\) 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: The rate of change in contamination -

\(\frac { dN }{ dt } =-\frac { 3150 }{ { t }^{ 4 } } -220\\ \\ N(t)\quad =\quad \int { (-\frac { 3150 }{ { t }^{ 4 } } -220) } dt\\ \quad \quad \quad =\quad -\int { (\frac { 3150 }{ { t }^{ 4 } } +220) } dt\\ \quad \quad \quad =\quad -3150\frac { { t }^{ -4+1 } }{ -4+1 } -220t+C\\ \quad \quad \quad =\quad \frac { 1050 }{ { t }^{ 3 } } -220t+C\)

Since, the level of contamination after 1 day was 6530 bacteria per cubic centimeter

Hence, applying t=1 on the definition of N(t),

\(N(t)\quad =\quad \frac { 1050 }{ { t }^{ 3 } } -220t+C\\ N(1)\quad =\quad 6530\quad =\quad \frac { 1050 }{ { 1 }^{ 3 } } -220.1+C\\ =>\quad C\quad =\quad 6530-1050+220\\ \quad \quad \quad \quad =\quad 5700\\ Hence,\quad N(t)\quad =\quad \frac { 1050 }{ { t }^{ 3 } } -220t+5700\)

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

Ans:

\(\int _{ 4.5 }^{ 8.5 }{ (2x-9)dx } \\ F(x)={ x }^{ 2 }-9x+C\\ Area\quad =\quad 8.5*8.5-9*8.5+C-4.5*4.5+9*4.5-C\\ \quad \quad \quad \quad =72.25-76.5-20.25+40.5\\ \quad \quad \quad \quad =16\)

#Define integrand function
integrand <- function(x)
  {
    2*x-9
  }

#Integrate the function from 4.5 to 8.5
integrate(integrand, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

The total area of the red rectangles are 16.

  1. Find the area of the region bounded by the graphs of the given equations -

\(y={ x }^{ 2 }-2x-2,\quad y=x+2\)

Enter your answer below.

Ans: Let’s first plot the graphs based on two equations-

#plot the graphs
curve ((x^2)-2*x-2, -5, 5)
curve (x+2, -5, 5, add=TRUE, col="blue")

Then, let’s find the points of intersection between the curves. At these points, the equations are supposed to be equal to each other -

\(y={ x }^{ 2 }-2x-2,\quad y=x+2\\ { x }^{ 2 }-2x-2\quad =\quad x+2\\ =>\quad { x }^{ 2 }-3x-4\quad =\quad 0\\ =>\quad (x-4)(x+1)\quad =\quad 0\)

#find the intersection pts by using the rootSolve package
intersect<- function (x){(x^2)-2*x-2-x-2}
library("rootSolve")
uniroot.all(intersect, lower=-5, upper=5)
FALSE [1] -1  4

To find the area of intersection between the two curves we need to perform definite integral between the points of intersection for the difference of the two functions -

\(F(x)=\int _{ }^{ }{ (x+2)-({ x }^{ 2 }-2x-2)dx } \\ =\quad \int _{ }^{ }{ (-{ x }^{ 2 }+3x+4)dx } \\ =\quad -\frac { { x }^{ 3 } }{ 3 } +\frac { 3 }{ 2 } { x }^{ 2 }+4x+C\\ F(4)-F(-1)\quad =\quad -\frac { (4)*(4)*(4) }{ 3 } +\frac { 3 }{ 2 } *(4)*(4)+4*(4)-\frac { (-1)*(-1)*(-1) }{ 3 } -\frac { 3 }{ 2 } *(-1)*(-1)-4*(-1)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 64 }{ 3 } +24+16+\frac { 1 }{ 3 } -\frac { 3 }{ 2 } +4\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 44-\frac { 63 }{ 3 } -\frac { 3 }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 264-126-9 }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 129 }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 21.5\quad\)

#integrate the function (top - bottom) from -1 to 4
integrand <- function(x)
  {
  (x+2) - ((x^2)-2*x-2)
  }
integrate(integrand, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13

The area of the region bounded is approximately 21.

  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.

Ans: Let us assume n be the number of flat irons to order to minimize the inventory cost. Also, let’s assume over the next year average no. of flat irons to be stored is n/2 since the orders will be spread across the next 12 months.

Yearly storage cost = Storage cost per flat iron x Average number of flat irons stored = 3.75*(n/2) = 1.875n

Yearly ordering cost = Cost of each order × Number of orders = 8.25*110/n = 907.5/n

Inventory cost = Yearly storage cost+Yearly ordering cost = 1.875n+907.5/n=f(n)

Now we will find the minimized value, differentiate and solve at 0:

\(f'(n)=1.875-\frac { 907.5 }{ { n }^{ 2 } } \\ f'(n)=0\\ =>1.875-\frac { 907.5 }{ { n }^{ 2 } } =0\\ =>\quad 1.875{ n }^{ 2 }=907.5\\ =>\quad { n }^{ 2 }=\frac { 907.5 }{ 1.875 } \\ =>\quad n=\sqrt { \frac { 907.5 }{ 1.875 } } \\ =>\quad n=\sqrt { 484 } \\ =>\quad n=22\)

As per the observation there has to be 22 flat irons, so there should be 110/22=5 orders.

  1. Use integration by parts to solve the integral below - \(\int { \ln { (9x).{ x }^{ 6 }dx } }\)

Ans: By Integration By Parts rule -

\(\int { udv } \quad =\quad uv-\int { vdu }\)

Let’s assume, \(u=\ln { (9x),\quad dv={ x }^{ 6 }dx } \\ =>\quad du=\frac { 1 }{ x } dx,\quad v=\frac { { x }^{ 7 } }{ 7 }\)

\(\int { \ln { (9x).{ x }^{ 6 }dx } } =\quad \frac { { x }^{ 7 } }{ 7 } .\ln { (9x) } -\int { \frac { { x }^{ 7 } }{ 7 } } .\frac { 1 }{ x } dx\\ \qquad \qquad \qquad =\quad \frac { { x }^{ 7 } }{ 7 } .\ln { (9x) } -\frac { 1 }{ 7 } \int { { x }^{ 6 }dx } \\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { { x }^{ 7 } }{ 7 } .\ln { (9x) } -\frac { { x }^{ 7 } }{ 49 } +C\\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { { x }^{ 7 } }{ 49 } (7\ln { (9x) } -1)+C\)

  1. Determine whether f(x) is a probability density function on the interval \(\left[ 1,{ e }^{ 6 } \right]\). If not, determine the value of the definite integral- \(f(x)=\frac { 1 }{ 6x }\)

Ans: In general, to determine the probability that X is in any subset A of the real numbers, we simply add up the values of Probability Density Function, p(x) in the subset. By “add up,” essentially means integrate the function p(x) over the set A. The probability that X is in A is precisely -

\(Pr(x\in A)\quad =\quad \int _{ A }^{ }{ p(x)dx }\)

For a function p(x) to be a probability density function, it must satisfy two conditions. It must be non-negative, so the that integral above is always non-negative, and it must integrate to one, so that the probability of X being something is one:

\(p(x)\ge 0\quad for\quad all\quad x\\ \int { p(x)dx } =1,\)

\(\int _{ 1 }^{ { e }^{ 6 } }{ f(x) } dx\quad =\quad \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } } dx\\ \qquad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 6 } (\ln { { e }^{ 6 } } -\ln { 1 } )\\ \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1-0\quad =\quad 1\)

Since in the interval \(\left[ 1,{ e }^{ 6 } \right]\), f(x) is always positive and the integral as calculated above is 1, we can conclude that f(x) is a probablity density function.