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

Let $u = -7x$ or

$\cfrac { du }{ dx } = -7\Rrightarrow$ ${ dx } = \cfrac { -1 du}{ 7 }$

Hence, ${\int 4e^{-7x}dx}$ $\Rrightarrow\cfrac {-4}{7}{\int e^{u}du}$

Now solving: ${\int e^{u}du}$ $\Rrightarrow e^u$ According to exponential rule.

Substituting this back to equation above ${\int 4e^{-7x}dx}$ $\Rrightarrow \cfrac {-4}{7}{\int e^{u}du}$ $\Rrightarrow \cfrac {-4}{7}e^u$

Substituting back the value of u:- $\cfrac {-4}{7}e^u$ $\Rrightarrow \cfrac {-4}{7}e^{-7x}$ $\Rrightarrow (-\cfrac {4}{7}e^{-7x} + C)$

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of $\color{red} {\cfrac { dN }{ dt } = -\cfrac { 3150 }{ t^4 } - 220}$ 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.

${\cfrac { dN }{ dt } = -\cfrac { 3150 }{ t^4 } - 220}\Rrightarrow$ $N^{'}(t) = \int {-\cfrac { 3150 }{ t^4 } - 220}. dt \Rrightarrow$ $\cfrac {1050} {t^3} -220t + C$

Since, $N(1) = 6530$ The Above Equation can be solved for value of C

6530 = $\cfrac {1050} {1^3} -220(1) + C$

$C = 6530-1050+220$

$C = 5700$

We can not replace the value of C to calculate $N^{'}(t)$ $=\cfrac {1050} {t^3} -220t +5700$

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

cal <- function(x)
  {
  (2*x)-9 
  }

integrate(cal, lower = 4.5, upper = 8.5)

## 16 with absolute error < 1.8e-13

4. Find the area of the region bounded by the graphs of the given equations. $\color{red} {y = x^2 - 2x - 2, y = x + 2}$ Enter your answer below.

curve ((x^2)-2*x-2, -10, 10)
curve (x+2, -10, 10, add=TRUE)

findintersect<- function (x){(x^2)-2*x-2-x-2}
library("rootSolve")
uniroot.all(findintersect, lower=-10, upper=10)

## [1] -1  4

area <- function(x)
  {
  (x+2) - ((x^2)-2*x-2)
  }
integrate(area, lower = -1, upper = 4)

## 20.83333 with absolute error < 2.3e-13

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.

Total number of Irons to sell = 110. Let x be a number of flat irons to order.

$Total Yearly storage cost=Storage cost×Average number of irons stored =$ $3.75× \cfrac{x}{2}=1.875x$

$Total Yearly ordering cost=Cost of each order×Number of orders =$ $8.25×\cfrac{110}{x}=\cfrac{907.5}{x}$

$Total Inventory cost=Yearly storage cost+Yearly ordering cost =$$1.875x+\cfrac{907.5}{x}=f(x)$

To find the minimal value, we calculate derivative and solve at 0 the above equation becomes:

$f^{'}(x) = 1.875-\cfrac{907.5}{x^2} = 0 \Rrightarrow$ $1.875 = \cfrac{907.5}{x^2} \Rrightarrow x = \sqrt {\cfrac{907.5}{1.875}}$

Solving above gives $x = 22$ Lots

Orders is calculated by:- number of irons to $\cfrac{Iron Sales}{Lots}$ $\Rrightarrow \cfrac{110}{22} = 5$ Orders

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

Integrate by parts formula tells us: $\int fg^{'} = fg - \int f^{'}g$

Let $f = ln(9x)$ and $g^{'} = x^6$ or $f^{'} = \cfrac {1}{x}$ and $g = \cfrac {x^7}{7}$

Substituting into above equation $\cfrac {x^7 ln(9x)}{7} - \int \cfrac{x^6}{7}dx$ $\Rrightarrow$ $\cfrac {x^7 ln(9x)}{7} - \cfrac {1}{7}\int {x^6}dx$

$\Rrightarrow$ $\cfrac {x^7 ln(9x)}{7} - \cfrac {1}{7} \times \cfrac{x^7}{7}$ $\Rrightarrow$ $\cfrac {x^7 ln(9x)}{7} -\cfrac{x^7}{49} + C$

Hence,

$x^7 \cfrac {(7ln(9x) -1)}{49} + C$

7. Determine whether f(x) is a probability density function on the interval $\color{red} {[1, e^6]}$ . If not, determine the value of the definite integral. $\color{red} {f(x) = \cfrac { 1 }{ 6x }}$

probdens <- function(x) {1/(6 * x)}
all.equal(1, integrate(probdens, lower = 1, upper = exp(6))$value)

## [1] TRUE

We can see from above the definite integral of the function is 1 and f(x) is a probability density function.

data-605-week14-Assignment

Ashish Kumar

04/28/2020

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

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

4. Find the area of the region bounded by the graphs of the given equations. \(\color{red} {y = x^2 - 2x - 2, y = x + 2}\) Enter your answer below.

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.

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

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