HW13 - Calculus

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

F(x) = $\int{4e^{-7x} dx}$ .

Let $u = -7x$ .
Then $du = -7dx$ , so $dx = \frac{du}{-7}$ .

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

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.

\[\begin{aligned} N(t)-N(0) &=\int\limits_0^t {\left(\frac{-3150}{s^{4}}-220 \right)ds}\\ &=\int\limits_0^t {\left(-3150{s^{-4}}-220 \right)ds}\\ &=\int\limits_0^t-3150\cdot s^{-4}ds - \int\limits_0^s 220ds\\ &=-3150\int\limits_0^t s^{-4}ds - 220\int\limits_0^t ds\\ &=\left[ -3150 \frac{-1}{3} s^{-3} - 220s \right]_{s=0}^{s=t}\\ &=\frac{3150}{3} t^{-3} - 220t -0+0\\ N(t)-N(0)&=\frac{1050}{ t^{3}} - 220t \\ \end{aligned}\]

We are given an initial condition: $N(t=1)=6530$ .

So, \[\begin{aligned} N(1)-N(0)&=\frac{1050}{ 1^{3}} - 220 \cdot 1 = 1050 - 220 = 830 \\ 6530-N(0)&=830\\ N(0) &= 6530 - 830 = 5700 \end{aligned}\]

Therefore, the solution is $N(t)= \frac{1050}{t^{3}}-220t+5700$ .

Note that the solution is not defined at $t=0$ .
Note also that the first term will become asymptotically zero, so the reduction in bacteria will soon become linear at -220 per cubic centimeter per day. At day 26, the amount of bacteria would become negative, which is of course not feasible …

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

By inspection, we can see that the area is 1+3+5+7 = 16 .

Using calculus, we compute as follows:

\[\begin{aligned} \int\limits_{4.5}^{8.5}{(2x-9)dx} &=[x^2-9x]_{x=\frac{9}{2}}^{x=\frac{17}{2}} \\ &=\left(\frac{17}{2}\right)^2-\frac{9\cdot17}{2} - \left(\frac{9}{2}\right)^2+\frac{9\cdot9}{2} \\ &= \frac{289-81}{4} - \frac{153-81}{2} \\ &= \frac{208}{4} - \frac{72}{2} \\ &= 52 - 36 \\ &= 16 \end{aligned}\]

4. Find the area of the region bounded by the graphs of the given equations: $y = x^2 -2x - 2$, $y = x + 2$

First, let’s determine where these curves intersect:
\[\begin{aligned} x^2 -2x - 2 &= x + 2 \\ x^2 -3x - 4 &= 0 \\ (x+1)(x-4) &= 0 \\ x &\in\{-1,4\} \end{aligned}\]

y1 <- function(x) x^2 - 2*x -2
y2 <- function(x) x + 2
curve(expr = y1(x), from = -1, to = 4, col="red")
curve(expr = y2(x), from = -1, to = 4, col="blue", add=TRUE)
title(main = "Computing the area between two curves")

So, we clearly want to compute the following quantity:

\[\begin{aligned} \int\limits_{-1}^4 {(x + 2) - ( x^2 - 2x -2)} &= \int\limits_{-1}^4 { (-x^2 + 3x +4)}dx \\ &= \left. -\frac{x^3}{3}+\frac{3x^2}{2}+4x \right]_{-1}^4 \\ &= -\frac{4^3}{3}+\frac{3\cdot 4^2}{2}+4 \cdot 4 - \left( -\frac{(-1)^3}{3}+\frac{3\cdot(-1)^2}{2}+4\cdot(-1) \right) \\ &= -\frac{64}{3}+\frac{48}{2}+16 - \left( \frac{1}{3}+\frac{3}{2}-4 \right) \\ &= -\frac{65}{3}+\frac{45}{2}+20 \\ &= -\frac{130}{6}+\frac{135}{6}+20 \\ &= 20+\frac{5}{6} \\ &= 20.8333333 \end{aligned}\]

We can double check using the “integrate” function:

integrate(y2, lower = -1, upper = 4)$value - integrate(y1, lower = -1, upper = 4)$value

## [1] 20.8333333

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.

We will make the following assumptions:

We will order 110 flat irons over the next year, and assume we will sell each one at a constant rate, i.e., one iron sold every $\frac{365}{110} \approx 3.32$ days.
We are not concerned with the purchase or sales price for the irons, as no information is provided.
During the year, we will place $numOrders$ orders, where each order will be placed for $lotSize = \frac{110}{numOrders}$ irons .
This means that the fixed cost associated with placing the orders will be $8.25 * numOrders$ dollars .
On average, the number of irons in storage will be $\frac{lotsize}{2}$ .
The cost associated with maintaining such storage (over the year) will be $3.75 * \frac{lotSize}{2}$ dollars.

Thus, the cost function that we seek to minimize is:

\[ \begin{aligned} inventoryCost &= 8.25 * numOrders + 3.75 * \frac{lotSize}{2} \\ &= 8.25 * numOrders + 3.75 * \frac{110}{2\cdot numOrders} \\ &= 8.25 * numOrders + \frac{3.75 * 55}{numOrders} \\ inventoryCost &= 8.25 * numOrders + \frac{206.25}{numOrders} \end{aligned}\]

To find the extremum, $\frac{d(inventoryCost)}{d(numOrders)} = 8.25 - \frac{206.25}{numOrders^2} = 0$
So, $8.25 = \frac{206.25}{numOrders^2}$ , or $(numOrders)^2 = \frac{206.25}{8.25}=25$ .

Therefore, $numOrders = 5$ and $lotSize = \frac{110}{numOrders} = \frac{110}{5} = 22$ .

Therefore, we should place 5 orders each year, with 22 flatirons in each order.

The cost associated with submitting the orders will be $8.25 * 5 = 41.25$ dollars,
and the cost associated with holding an average of $\frac{22}{2}=11$ irons in inventory will also be $3.75 *11 = 41.25$ dollars,
for a total inventory cost of $ 82.50 .

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

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

The formula for integration by parts:

\[\begin{aligned} \int u \cdot dv &= u \cdot v - \int v \cdot du \\ \int {ln( 9x ) \cdot x^6dx} &= ln(9x) \cdot \frac{x^7}{7} - \int \frac{x^7}{7} \frac{dx}{x} \\ &= \frac{x^7 \cdot ln(9x)}{7} - \int \frac{x^6dx}{7} \\ &= \frac{x^7 \cdot ln(9x)}{7} - \frac{x^7dx}{7 \cdot 7} + C \\ &= \frac{x^7}{7}\left[ln(9x)+\frac{1}{7}\right] + C \end{aligned}\]

7. Determine whether $f ( x ) = \frac{1}{6x}$ is a probability density function on the interval $[1, e^6]$ .

If not, determine the value of the definite integral $\int\limits_1^{e^6}{\frac{1}{6x}dx}$ .

For the function to be a PDF on the interval,

the value must always be non-negative, and
the definite integral needs to equal 1.

For the first requirement:
$f ( x ) = \frac{1}{6x}>0$ for $x>0$, so this holds for $x \in [1, e^6]$ .

For the second requirement: \[\begin{aligned} \int\limits_1^{e^6}{\frac{1}{6x}dx} &= \left. \frac{ln(x)}{6} \right]_1^{e^6} \\ &= \frac{ln(e^6)-ln(1)}{6} \\ &= \frac{6-0}{6} \\ &= 1 \end{aligned} \]

Therefore, the function is a PDF on the interval.

605-HW13-Calculus

Michael Y.

November 24, 2019

HW13 - Calculus

1. Use integration by substitution to solve the integral \(\int{4e^{-7x} dx}\) .

2. Biologists are treating a pond contaminated with bacteria.

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

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

5. A beauty supply store expects to sell 110 flat irons during the next year.

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

7. Determine whether \(f ( x ) = \frac{1}{6x}\) is a probability density function on the interval \([1, e^6]\) .