# 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.