Exercise 1

Use the integration by subsitution to solve the integral below:

\[\int 4e^{-7x}dx\] According to https://www.mathsisfun.com/calculus/integration-by-substition.html Integration by Substitution aka u-substituion is a method to find an integral.

\[\int 4e^{-7x}dx =\int 4e^u x -(\frac{1}{7})du\]

\[=\frac{-4}{7} \int e^udu\] \[=\frac{-4}{7}e^{-7x} + C\]

Exercise 2

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.

\[N(t) = \frac{1050}{t^3} - 220t + C\] \[N(1) = 6530\] \[\frac{1050}{1^3} - 220 x 1 + C = 6530\]

cat("Solution for C","\n")
## Solution for C
6530 - 1050 + 220
## [1] 5700

Exercise 3

Find the total area of the red rectangles in the figure below (figure not listed), where the equation of the lines is

\[f(x) = 2x - 9\]

Data points are based off assignment graph:

cat("Area 1","\n")
## Area 1
(a1 <- (5.5 - 4.5) * 1)
## [1] 1
cat("\n","Area 2","\n")
## 
##  Area 2
(a2 <- (6.5 - 5.5) * 3)
## [1] 3
cat("\n","Area 3","\n")
## 
##  Area 3
(a3 <- (7.5 - 6.5) * 5)
## [1] 5
cat("\n","Area 4","\n")
## 
##  Area 4
(a4 <- (8.5 - 7.5) * 7)
## [1] 7
cat("\n","Total Area","\n")
## 
##  Total Area
(a1 + a2 + a3 + a4)
## [1] 16

\[ Area = \int_{4.5}^{8.5} 2x - 9dx\] \[ Area = [8.5^2 - 9 * 8.5] - [4.5^2 - 9 * 4.5]\]

cat("Total Area of the Red Rectangles","\n")
## Total Area of the Red Rectangles
(area <- (8.5**2 - 9 * 8.5) - (4.5**2 - 9 * 4.5))
## [1] 16

Exercise 4

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

\[y = x^2 - 2x - 2, y = x + 2\]

Exercise 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.

\[= 8.25 + \frac{3.75x}{2}\]

\[ 8.25 + \frac{206.25}{x}\]

\[ 8.25 - \frac{206.25}{x^2}\]

cat("Number of Orders Per Year","\n")
## Number of Orders Per Year
(orders <- sqrt(206.25/8.25))
## [1] 5

Exercise 6

Use integration by parts to solve the integral below.

\[\int ln(9x) . x^6dx\] #Formula \[\int v\frac{du}{dx} dx = uv - \int u\frac{dv}{dx} dx\] \[ln(9x)\frac{1}{7}x^7 - \int\frac{1}{7}x^7\frac{1}{x}dx\] \[ln(9x)\frac{x^7}{7} - \frac{x^7}{49}-C\] \[v = \frac{1}{7}x^7\]

\[du = \frac{1}{x}\]

Exercise 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.

\[\int_{1}^{e^6} \frac{1}{6x} dx = \frac{1}{6}ln(x)|\prescript{e^6}{1}\] \[\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1)\]

\[\frac{1}{6} x 6 - \frac{1}{6} x 0\]

Based on the calculation the solution is 1