Question 1

Use integration by substitution to solve the integral below.

Sub for dx:
\(∫4e^(-7x) dx\)
\(u = 7x\)
\(\frac{du}{dx} = -7\)
\(du = -7dx\)
\(dx = \frac{du}{-7}\)

Plug in:
\(∫4e^(-7x) dx\) -> \(∫4e^u \frac{du}{-7}\)
Take out integers:
\(-\frac{4}{7}∫e^u du\)
\(-\frac{4}{7}e^u + C\)
Bring back u:
\(-\frac{4}{7}e^(-7x) + C\)

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.

Get N(t):
\(\frac{dN}{dt} = -\frac{3150}{t^4} - 220\)
\(N(t) = ∫\frac{-3150}{t^4} - 220 dt\)
\(N(t) = \frac{-3150}{-3t^3} - 220t\)
\(N(t) = \frac{1050}{t^3} - 220t + c\)
Now to solve for c against the outlined variables:
\(N(t) = 6530 & t = 1 day\)
\(6530 = \frac{1050}{1^3} - 220(1) + c\)
\(6530 = 1050 - 220 + c\)
\(6530 = 830 + c\)
\(5700 = c\)

So, our 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.

hw13_q3
The graph shows x at 4.5 and 8.5
\(∫2x - 9 dx\) |8.5 & 4.5
\(x^2 - 9x\) |8.5 & 4.5
\(Area = (x^2 − 9x)|8.5 - (x^2 - 9x)|4.5\)
\(= (8.5^2 − 9(8.5)) - (4.5^2 − 9(4.5))\)
\(= (72.25 − 76.5) - (20.25 − 40.5)\)
\(= (−4.25) − (−20.25)\)
Area = 16

Question 4

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

First, we need to find the x bounds by setting the equations equal to each other and solve for x - the intersections.
\(x^2 - 2x - 2 = x + 2\)
\(x^2 - 3x - 4 = 0\)
\((x - 4)(x + 1)\)
\(x = 4, -1\)

Now, we solve it like question 3.
\(∫(x + 2) - (x^2 - 2x - 2) dx\) |4 & -1
\(∫-x^2 + 3x + 4 dx\) |4 & -1
\(\frac{x^3}{3} + \frac{3x^2}{2} + 4x\) |4 & -1
\(4(\frac{x^3}{3} + \frac{3x^2}{2} + 4x) - -1(\frac{x^3}{3} + \frac{3x^2}{2} + 4x)\)
\((\frac{4^3}{3} + \frac{(3*4)^2}{2} + (4*4)) - (\frac{-1^3}{3} + \frac{(3*-1)^2}{2} + (4*-1))\)
\((\frac{64}{3} + \frac{144}{2} + (16)) - (\frac{-1}{3} + \frac{9}{2} + (-4))\)
\((\frac{64}{3} + 72 + 16) - (\frac{-1}{3} + 4.5 - 4)\)
\((\frac{64}{3} + 88) - (\frac{-1}{3} + .5)\)
\((\frac{328}{3}) - (\frac{1}{6})\)
\((\frac{655}{6})\) or 109.166667

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.

To solve this, I will use the economic order quantity formula:
\(EOQ = \sqrt(\frac{2SD}{H})\) where
S = Fixed cost to purchase = 8.25
D = annual demand = 110
H = inventory carrying cost = 3.75

\(\sqrt(\frac{2*8.25*110}{3.75})\)
\(\sqrt(\frac{1815}{3.75})\)
\(\sqrt(484) = 22\)

So the lot size is 22.
To fix the min number of orders, we can look at how many lots we would need to sell to break even:
22 * x = 110 -> x = 5
Our min number of orders is 5.

Question 6

Use integration by parts to solve the integral below.

\(∫ln(9x)·x^6 dx\)

Formula: \(∫i(x)j'(x) dx = ij - ∫i'(x)j(x) dx\)
We need some variables first:
\(i = ln(9x)\)
\(j' = x^6\)
\(i' = \frac{1}{x}\)
\(j = \frac{x^7}{7}\)

\(= ln(9x)*\frac{x^7}{7} - (∫\frac{1}{x}*\frac{x^7}{7} dx)\)
\(= \frac{x^7ln(9x)}{7} - (∫\frac{x^6}{7} dx)\)
\(= \frac{x^7ln(9x)}{7} - (\frac{1}{7}∫x^6 dx)\)
\(= \frac{x^7ln(9x)}{7} - (\frac{1}{7}*\frac{x^7}{7})\)
\(= \frac{x^7ln(9x)}{7} - \frac{x^7}{49}\)
\(= \frac{x^7ln(9x)}{7} - \frac{x^7}{49} + c\)

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

To do this, we check if the probability density function solves to 1:
\(∫\frac{1}{6x} dx\) | \(e^6, 1\)
\(\frac{1}{6}∫\frac{1}{x} dx\) | \(e^6, 1\)
\(\frac{1}{6}*ln(x)\) | \(e^6, 1\)
\(\frac{ln(e^6)}{6} - \frac{ln(1)}{6}\)
\(\frac{6ln(e)}{6} - \frac{0}{6}\)
\(\frac{6*1}{6} - 0\)
\(\frac{6}{6} = 1\)
So, f(x) is a probability density function on the interval \([1,e^6]\)