Question 1

Use integration by substitution to solve the integral below.

\[ \int{4e^{-7x}dx}\]

we have

\[ \int{4e^{-7x}dx} = 4 \int{e^{-7x}dx}\]

Let u = -7x

\[ du = -7dx \Leftrightarrow \frac{du}{-7}=dx \Leftrightarrow dx = -\frac{1}{7}du\]

Therefore:

\[ \begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} \\ &= \int{\frac{-4}{7}e^u du} \\ &= \frac{-4}{7}e^u+ C \\ &= -\frac{4}{7}e^{-7x}+ C \end{split} \]

Question2

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.

\[ \begin{split} \int{-\frac{3150}{t^4}-220 dt} &= -220t-3150\frac{1}{-3t^3} \\ &= -220t+\frac{1050}{t^3} + C \end{split} \]

Solving for \(C\) given that \(N(1) = 6530\)

\[ \begin{split} 6530 &= -220(1)+\frac{1050}{(1)^3} + C \\ \Leftrightarrow 6530 &= -220 + 1050 + C \\ \Leftrightarrow 6530 &= 830 + C \\ \Leftrightarrow 5700 &= C \end{split} \]

The function is \(N(t)=\frac{1050}{t^3}-220t+5700\)

Question 3

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

The area under the line is a triangle from 4.5 to 8.5. The height is found by pluging 8.5 into the function.

\[f(8.5)=2(8.5)-9=17-9=8\]

The area of the triangle is 1/2 base x height so

\[ \begin{split} \frac{1}{2}(8.5 - 4.5) \times 8 &= \frac{1}{2}(4) \times 8 \\ &= 2 \times 8 \\ &= 16 \end{split} \]

Our area is therefore \(\textbf{16}\).

If we use integration, then:

\[ \begin{split} \text{ Area } &= \int_{4.5}^{8.5}{(2x-9)dx} \\ &= \left.x^2-9x\right|_{4.5}^{8.5} \\ &=[8.5^2 - 9(8.5)] - [4.5^2 - 9(4.5)] \\ &=-4.25 + 20.25 \\ &= 16 \end{split} \]

Let’s check it with R

myfunc <- function(x){
  2*x - 9
}
integrate(myfunc, 4.5, 8.5)
## 16 with absolute error < 1.8e-13

Question 4

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

\[ \begin{split} f(x) &= x^2-2x-2 \\ \text{ and } g(x) &= x+2 \end{split} \]

First we will visualize the two lines. The first is blue and the second is red.

The two points where the lines intersect are \[x^2-2x-2 = x+2 \Rightarrow x^2-3x -4 = 0 \Rightarrow (x+1)(x-4) = 0 \Rightarrow x = -1 \text{ or } x = 4\]

Our area of interest is over the interval [-1, 4] and therefore:\[\text{ Area } = \int_{-1}^4{(x+2)-(x^2-2x-2)} \]

myFunc4 <- function(x){
  (x+2)-(x^2 - 2*x - 2)
}
integrate(myFunc4, -1, 4)
20.83333 with absolute error < 2.3e-13

Our Area over the interval [-1,4] is therefore \(\textbf{20.83}\)

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.

Let \(x\) = number of flat irons to order.

\(Yearly\ storage\ cost = {Storage\ cost\ per\ iron} \times {Average\ number\ of\ irons\ stored} = 3.75 \times x/2 = 1.875x\)

\(Yearly\ ordering\ cost = {Cost\ of\ each\ order} \times {Number\ of\ orders} = 8.25 \times 110/x = 907.5/x\)

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

To minimize inventory cost, we need to differentiate and then solve for this differentiation when it’s \(0\):

\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \\ \Rightarrow 1.875-\frac{907.5}{x^2} &= 0 \\ \Rightarrow 1.875&= \frac{907.5}{x^2} \\ \Rightarrow 1.875x^2&= 907.5 \\ \Rightarrow x^2&= \frac{907.5}{1.875} \\ \Rightarrow x&= \sqrt{\frac{907.5}{1.875}} \\ \Rightarrow x&=\sqrt{484} \\ \Rightarrow x&=22 \end{split} \]

Each order should contain \(22\) flat irons, so there should be \(110/22=5\) orders.

So: \[ \begin{split} \text {Lot size} &= \textbf{22} \\ \text {Number of Orders per year} &= \textbf{5} \end{split} \]

Question 6

Use integration by parts to solve the integral below.

\[\int{ln(9x) x^6 dx}\]

The formula for integration by parts is:

\[\int{f(x)g'(x)}=f(x)g(x)-\int{f'(x)g(x)}\]

Let \(f(x) = ln(9x)\) and let \(g'(x) = x^6\).

Therefore:

\(f'(x)=\frac{1}{x}\) and \(g(x) = \int{x^6}=\frac{x^7}{7}\).

Substituting, we’ll get:

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

Evaluating the sub-integral

\[\int{\frac{1}{x}\frac{x^7}{7}} = \int{\frac{x^6}{7}} = \frac{x^7}{49} + C\] Therefore: \[ \begin{split} \int{ln(9x) x^6 dx} &= ln(9x)\frac{x^7}{7}- \frac{x^7}{49} + C\\ &= 7ln(9x)\frac{x^7}{49}- \frac{x^7}{49} + C\\ &= \frac{x^7}{49}(7ln(9x)-1) + C \end{split} \]

Question 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.

\[f(x)=\frac{1}{6x}\]

If f(x) is a probability density function, then the area under the function’s curve within the interval should equal 1.i.e.: \[F(x) = \int_1^{e^6}{\frac{1}{6x}dx}=1\]

Let’s verify it.

\[ \begin{split} \int_1^{e^6}{\frac{1}{6x}dx} &= \frac{1}{6}ln(x) |_1^{e^6} \\ &= \frac{1}{6}ln(e^6) - \frac{1}{6}ln(1) \\ &=\frac{1}{6}\times6 - \frac{1}{6}\times 0 \\ &= 1 - 0 \\ &= 1 \end{split} \]

The area under the curve over the interval \([1,e^6]\) is indeed \(\textbf{1}\), therefore f(x) is a probability density function.