Question 1

Use integration by substitution to solve the integral below. \[\int{4e^{-7x}dx}\]

Solution 1

Let \(u=-7x\), => \(du = -7dx\). \(dx = -\frac{du}{7}\) Therefore, \[ \begin{split} \int{4e^{-7x}dx} = \int{4e^{u}\left ( \frac{-du}{7} \right )} \\ = -\frac{4}{7}\int{e^u du} \\ = \frac{-4}{7}e^u + k \\ = -\frac{4}{7}e^{-7x} + k \end{split} \]

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.

Solution 2

\[ \begin{split} \frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220 \\ N(t) = \int{(\frac{-3150}{t^4}-220) dt} \\ = \frac{1050}{t^3}-220t + k \\ N(1) = 6530, therefore \\ N(t) = \frac{1050}{t^3}-220t+k \\ 6530 = \frac{1050}{1^3}-220\times 1 + k \\ 6530 - 1050 + 220 = k \\ k = 5700 \end{split} \]

If \(N(1)= 6530\), then the level of contamination can be estimated by the function below:

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

Question 3

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

Solution 3

Each square in the graph has an area of \(1\) unit. Each rectangle in the graph has a width of \(1\) unit as well. Counting the height of each rectangle marked in yellow gives the areas as:
\(Area =1+3+5+7=16\) sq units.

Question 4

Find the area of the region bounded by the graphs of the given equations.
\(y = x^2 - 2x-2\), \(y = x + 2\)

Solution 4

To find the area of the region bounded by the two equations, we first need to find the points where the two equations meet by setting the equations equal to one another.

\[ x^{2} − 2x − 2 = x + 2 \\ x^{2}−2x-x-2−2 = 0 \\ x^{2}−3x-4 = 0 \\ (x-4)(x+1) = 0 \\ x = -1, 4 \] Therefore upper limit of the integral is 4 while the lower limit is -1

\[ \int_{-1}^{4} [(x^{2}-2x-2)-(x+2)]dx \\ = \int_{-1}^{4}(x^{2}-3x-4)dx \\ = [\frac{x^{3}}{3} - \frac{3x^{2}}{2} - 4x]_{-1}^{4} \\ = (\frac{64}{3} - \frac{3*16}{2} - 16) - (\frac{-1}{3} - \frac{3}{2} + 4)\\ = -20.833 \]

Since area cannot be negative, the area of the region bounded by those two equations is about 20.833 sq units

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.

Solution 5

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

\[ Annual\ storage\ cost = {Storage\ cost\ per\ iron} \times {Average\ number\ of\ irons\ stored} = 3.75 \times x/2 = 1.875x \\ Annual\ order\ cost = {Cost\ of\ each\ order} \times {Number\ of\ orders} = 8.25 \times 110/x = 907.5/x \\ Inventory\ cost = Annual\ storage\ cost + Annual\ order\ cost = 1.875x+907.5/x = f(x) \]

To find the lot size that will minimized inventory costs, we find the first derivative of the inventory cost and find the value at x = 0 :

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

Hence, each order should contain \(22\) flat irons, and there should be \(110/22=5\) orders to minimize inventory costs.

Question 6

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

Solution 6

Using the integration by parts approach \[ Let\ u= ln(9x),\ then\ \frac{du}{dx}=\frac{1}{x} \\ Also,\ let\ \frac{dv}{dx}=x^6,\ therefore,\ v = \int{x^6 dx} = \frac{x^7}{7} \\ = \int{ln(9x).x^6dx} = ln(9x).\frac{x^7}{7} - \int{\frac{x^7}{7} \times \frac{dx}{x}} \\ = ln(9x) \times \frac{x^7}{7} - \int{\frac{x^6}{7}dx} \\ = ln(9x) \times \frac{x^7}{7} - \frac{x^7}{49} + k \\ = \frac{x^7}{49}(7ln(9x) - 1) + k \\ \]

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. \(f(x) = \frac{1}{6x}\)

Solution 7

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