Question_1

Use integration by substitution to solve the integral below.

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

\[=\int { \frac { 4\times -7 }{ -7 } } { e }^{ u }\quad du\] \[=\frac { -4 }{ 7 } { e }^{ -7x }\quad +\quad c\] where c is a constant.

Question_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.

\[\frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220\] To find \(N(t)\) we find the integration

\[N(t) = \int{(\frac{-3150}{t^4}-220) dt}\] \[N(t)= \frac{1050}{t^3}-220t+C\] back substitute with the \(N(t) = 6530\) where t = 1 to find C

\[6530= \frac{1050}{(1)^3}-220(1)+C\] \[5700= C\]

The contamination level could be presented by

\[N(t)= \frac{1050}{t^3}-220t+5700\]

Question_3

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

from the graph provided, we can see that each column have a width of 1 unit. The length can be found using the integration and substitute with the min and max value for the rectangle width.

eq = function(x) {
  return (2*x - 9)
}
area = integrate(eq, 4.5, 8.5)
print(area)
## 16 with absolute error < 1.8e-13

Question_4

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

\[y_1 = x_1^2 - 2x_1-2\] \[y_2 = x_2 + 2\]

y_1 = function(x) {
  return(x^2 - 2*x - 2)
}

y_2 = function(x) {
  return(x + 2)
}

plot(y_1, -3, 6, col = 'red')
plot(y_2, -3, 6, add=TRUE, col = 'blue')

We need to solve the two equations for x to get the intersection points, (the intersection points are the bounds for the integral)

\[x^{2}-2x-2=x+2\]

\[x^{2}-3x-4=0\] \[x=4 \quad or \quad x=-1\]

\[A=\int _{ { \, a } }^{ { \, b } }{ { \left( \begin{array}{c} { { upper } } \\ { { function } } \end{array} \right) -\left( \begin{array}{c} { { lower } } \\ { { function } } \end{array} \right) \, dx } } \]

# upper function is the y_2, lower function is the y_1
area_1 <- integrate(y_1, -1, 4)
area_2 <- integrate(y_2, -1, 4)
area = (area_2$value - area_1$value)
print(as.numeric(area))
## [1] 20.83333

Question_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.

#of orders per year = n lot size = 110/n cost = lot size * store cost + fixed cost * n cost = 110/n

assume half of the inventory are in stock

\[cost = \frac { 110 }{ 2n } \times 3.75\quad +\quad 8.25n\\ \] \[cost = \frac { 206.25 }{ n } \quad +\quad 8.25n\\ \] take first integral for n

\[\int { cost } \quad =\quad \int { \frac { 206.25 }{ n } } +\quad 8.25n\] Now we can get the number of orders per year that will minimize the cost which is zero

\[\quad -\frac { 206.25 }{ { n }^{ 2 } } \quad +\quad 8.25\quad =\quad 0 \]

Solve for n we get that \(n = 5\), \(lot size = 22\)

Question_6

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

\[=\frac { 1 }{ 7 } x^{ 7 }\times ln(9x)-\int { \frac { 1 }{ 7 } x^{ 7 }\times \frac { 1 }{ x } dx } \\ =\frac { 1 }{ 7 } x^{ 7 }\times ln(9x)-\int { \frac { 1 }{ 7 } x^{ 6 }dx } \\ =\frac { 7 }{ 49 } x^{ 7 }\times ln(9x)-\frac { 1 }{ 49 } x^{ 7 }+c\\ =\frac { 1 }{ 49 } x^{ 7 }(7ln(9x)-1)+c\\ \]

Question_7

Determine whether \(f(x)\) is a probability density function on the interval [1,e6]. If not, determine the value of the definite integral. \(f(x) = \frac{1}{6x}\)

\[\int _{ 1 }^{ e^{ 6 } } \frac { 1 }{ 6x } dx\quad =\quad \frac { 1 }{ 6 } ln(x)|_{ 1 }^{ e^{ 6 } }\\ =\frac { 1 }{ 6 } ln(e^{ 6 })\quad -\quad \frac { 1 }{ 6 } ln(1)\\ =\frac { 1 }{ 6 } \times 6\quad \\ =1\] The definite integral of the function on interval [1,e6] is 1. Additionally, if x>0, then f(x)>0, so for this interval f(x)>0. As long as f(x)=0 outside of the given interval, this satisfies PDF requirements and this function is a probability density function.