Question 1

Use integration by substitution to solve the integral below.

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

We can set \(u = -7x\). That means

\[ \begin{align} du &= -7 \ dx \\ \\ \frac{du}{-7} &= dx \end{align} \]


When we substitute \(u\) and \(dx\) back in the original function, we get

\[ \begin{align} &\int 4e^{u} \ \frac{du}{-7} &\qquad \text{simplify} \\ \\ &= \frac{-4}{7} \int e^{u} &\qquad \text{substitute } u \\ \\ & = \frac{-4}{7} e^{-7x} + C \\ \end{align} \]



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.

First, calculate the integral of both sides to determine \(N(t)\), or the level of contamination \(N\) at certain point in time \(t\).

\[ \begin{align} \int \frac{dN}{dt} &= \int \frac{-3150}{t^4} -220 \\ \\ N(t) &= \frac{3150}{3t^3} -220t + C \\ \\ N(t) &= \frac{1050}{t^3} -220t + C \\ \end{align} \]


Then, find the value of \(C\) using the given fact that \(N(1) = 6530\).

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


Finally, substitute \(C\) to get a general function \(N(t)\) for contamination.

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



Question 3

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

To solve this problem, find the areas of each of the rectangles and add them together.

\[ \begin{align} R_1 &= 1 \times 1\\ R_2 &= 1 \times 3 \\ R_3 &= 1 \times 5 \\ R_4 &= 1 \times 7 \\ \\ \text{Area} &= 1 + 3 + 5 + 7 = 16 \text{ square units} \end{align} \]

Confirm the solution by evaluating the integral of the function over the interval \(4.5\) to \(8.5\).

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



Question 4

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

\[ \begin{align} y_1 &= x^2 - 2x - 2\\ y_2 &= x + 2 \end{align} \]

Find where the functions intersect by setting them equal to one another and solving for \(x\).

\[ \begin{align} x+2 &= x^2-2x-2 \\ 0 &= x^2-3x-4 \\ 0 &= (x+1)(x-4) \\ \\ x &= -1 \ \text{ and } \ 4 \end{align} \]


Then, evaluate the integral of \(y_2 - y_1\) over the interval \(-1\) to \(4\) to find the area.

\[ \begin{align} \text{Area} &= \int_{-1}^{4} (x+2) \ dx \ - \int_{-1}^{4} (x^2-2x-2) \ dx \\ \\ \ &= \int_{-1}^{4} (-x^2+3x+4) \ dx \\ \\ \ &= \left. \frac{-x^3}{3} + \frac{3x^2}{2} + 4x \ \ \right|_{-1}^{4} \\ \\ \ &= [\frac{-4^3}{3} + \frac{3(4^2)}{2} + 4(4)] - [\frac{1^3}{3} + \frac{3(-1^2)}{2} + 4(-1)] \\ \\ \ &= \frac{-64}{3} + \frac{48}{2} + 16 - \frac{1}{3} + \frac{3}{2} + 4 \\ \\ \ &= \frac{-63}{3} + \frac{51}{2} + 20 \\ \\ \ &= 21.5 \text{ units squared}\\ \end{align} \]

(-63/3) + (45/2) + 20
## [1] 21.5


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.

First, define the equation to be minimized. Let \(x\) be the number of flat irons required per order.

\[ \begin{align} \text{Cost of storage} &= \frac{3.75x}{2} \\ \\ \text{Cost per order} &= 8.75 \times \frac{110}{x} \\ \\ \text{Inventory cost} &= (\frac{3.75x}{2}) \ + \ (8.75 \times \frac{110}{x}) \\ \\ f(x) &= 907.5x^{-1} + 1.875x \end{align} \]


Then, find the critical point by taking the derivative of the cost function, setting it equal to zero, and solving for x.

\[ \begin{align} f'(x) &= \frac{907.5}{x^2} - 1.875 \\ \\ 0 &= \frac{907.5}{x^2} - 1.875 \\ \\ 1.875x^2 &= 907.5 \\ \\ x^2 &= \frac{907.5}{1.875} = 484 \\ \\ x &= \sqrt{484} = 22 \text{ flat irons per order} \end{align} \]


Each order should contain 22 flat irons. That means the store should get at least \(\frac{110}{22} = 5 \text{ orders per year}\) to minimize inventory costs.



Question 6

Use integration by parts to solve the integral below.

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

The formula for integration by parts is:

\[ \int u \ dv = uv - \int v \ du \]


Let \(\ u = ln(9x)\ \) and \(\ dv = x^6\). Then, \(\ du = \frac{1}{x}\ \) and \(\ v = \frac{x^7}{7}\ \).


Subsitute in the formula and simplify:

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



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 \(1\).

\[ \begin{align} \int_{1}^{e^6} \frac{1}{6x} \ dx &= \left. \frac{1}{6} ln(x) \ \right|_{1}^{e^6} \\ \\ &= \frac{1}{6} \times ln(e^6) \ - \ \frac{1}{6} \times ln(1) \\ \\ &= \frac{1}{6} \times \frac{6}{1} \ - \ 0 \\ \\ &= 1 \end{align} \]


The area under the curve for the interval \([1, e^6]\) is 1, which means \(f(x)\) is a probability density function.