DATA605 - Homework 13

Week 13, Univariate & Multivariate Calculus: 23 Apr - 1 May

Exercise 1

Use integration by substitution to solve the integral below.

\(\int4e^{-7x}dx\)

Solution

\(\int4e^{-7x}dx\)

Define a new variable u as follows:

\(u = -7x => du = -7dx => dx = -\frac{1}{7}du\)

Now replace original integral in terms of \(u\) variable:

\(\int -\frac{4}{7}e^{u}du\) \(=-\frac{4}{7}\int e^{u}du\)

Now solving the integral we get:

\(=-\frac{4}{7}e^{u}\)

Undo substitution \(u=-7x\):

\(=-\frac{4}{7}e^{-7x}+C\)

Exercise 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

\(\frac {dN}{dt} = - \frac {3150}{t^4} - 220\)

Apply the anti-derivative (Integrate) \(\frac {dN}{dt}\) to obtain \(N( t )\):

\(\int (- \frac {3150}{t^4} - 220) dt\)

\(= -3150 \int \frac {1}{t^4} - 220 \int 1 dt\)

\(= -3150 (- \frac {1}{3t^3}) - 220 t\)

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

Before finding the final \(N( t )\), we need to calculate the value of the integration constant \(C\) from the last expression. We can also use the given information that after 1 day (t=1) the level of contamination was 6530 bacteria per cubic centimeter.

\(6530 = \frac {1050}{1^3} - 220 t + C\)

$ C = 6530 + 1270 = 7800$

Therefore,

\(N(t) = = \frac {1050}{t^3} - 220 t + 7800\)

Exercise 3

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

Exercise 3 red rectangles under a line

Solution

From the graph we can see that there are 4 red rectangles, whose areas can be calculated using the grid. The total area of the rectangles is:

\(Area_{rectangles} = 1 + 3 + 5 + 7 = 16 u^2\)

To find a more accurate estimation of the area under the curve by integrating the function over the interval \([4.5, 8.5\)

\(Area = \int_{4.5}^{8.5} (2x - 9) dx\)

Define the function f(x):

f <- function(x) {2*x - 9}

Integrate \(f( x ) = 2x - 9\) over the interval \([4.5, 8.5]\):

f_area <- integrate(f, lower = 4.5, upper = 8.5)
f_area_value <- f_area$value
f_area

#> 16 with absolute error < 1.8e-13

\(Area = \int_{4.5}^{8.5} (2x - 9) dx = {16}\)

Apparently, the rectangle estimation and the integration based estimations yield the same result in this case. It looks like the gaps under the curve left by the rectangles are compensated by the small areas of the rectangles over above the line. That would justify why the two approaches yield the same results.

Exercise 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.

Solution

Define the functions

f <- function(x) {x^2 - 2*x - 2}
g <- function(x) {x + 2}

Graph the functions

curve(f, from = -5, to = 6, col = 2)
curve(g, from = -5, to = 6, add = TRUE, col = 4)

g_area <- integrate(g, lower = -1, upper = 4)
g_area

#> 17.5 with absolute error < 1.9e-13

f_area <- integrate(f, lower = -1, upper = 4)
f_area

#> -3.333333 with absolute error < 1.2e-13

print(paste0('The area between the curves is ',(g_area$value - f_area$value), ' squared units.'))

#> [1] "The area between the curves is 20.8333333333333 squared units."

Exercise 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

Using the provided info a function that represents the inventory cost.

Let \(x\) be the number of flat irons to be ordered.

Let \(f(x)\) be the function that defines the inventory cost as follows:

f(X) = Yearly Storage Cost + Yearly Ordering Cost

where,

Yearly Storage Cost = Storage cost per iron * Average number of irons stored

\(= 3.75 * \frac{x}{2}\)

\(= 1.875 x\)

Yearly Ordering Cost = Cost per order * Number of orders per year

\(= 8.25 * \frac {110}{x}\)

\(= \frac {907.5}{x}\)

Therefore,

\(f(x) = 1.875 x + \frac {907.5}{x}\)

In order to find the lot size and the number of orders per year that will minimize inventory costs, find the derivative f’(x) and use it to find the minimum values.

\(f'(x) = 1.875 - \frac {1815}{2x^2}\)

Find the roots for \(f'(x)\)

\(f'(x) = 0\)

\(1.875 - \frac {1815}{2x^2} = 0\)

\(1.875 = \frac {1815}{2x^2}\)

\(1.875 = \frac {1815}{2x^2}\)

\((2)(x^2) (1.875) = 1815\)

\(x^2 = \frac {1815}{(2)(1.875)}\)

\(x^2 = 484\)

\(x = \sqrt{484} => x = -22\) or \(x = 22\)

Therefore, we can conclude that in order to minimize inventory costs, each order should contain 22 flat irons and there should be 110/22 = 5 orders per year.

Exercise 6

Use integration by parts to solve the integral below.

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

Solution

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

Integration by parts states:

\(\int f g' = f g - \int f' g\) \(~~~\)(Eq. A)

Define the functions \(f\) and \(g'\):

\(f = ln(9x)\)

\(g' = x^6\)

Calculate \(f'\) and \(g\):

\(f' = \frac{1}{x}\)

\(g = \int g' = \int x^6 = \frac{x^7}{7}\)

Replace the calculated values into equation (Eq. A)

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

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

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

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

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

\(= \frac {1}{7}(x^7)(ln(9x)) - \frac {1}{7} \frac {x^7}{7} + C\)

\(= \frac {1}{7}(x^7)(ln(9x) - \frac {1}{7}) + C\)

Exercise 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

For the function \(f(x)\) to be a PDF the following two conditions must be met:

(Condition 1) \(f(x) \geq 0, \forall x \in [1, e^6]\)

(Condition 2) \(\int_{1}^{e^6} f(x) = 1\)

Define the function \(f(x) = 1/6x\)

f <- function(x) {1/(6*x)}
e <- exp(1)

Test (Condition 1) by graphing the function over the interval \([1, e^6]\):

curve(f, from = 0, to = e^6)

From the graph above, we can see that \(f(x) \geq 0, \forall x \in [1, e^6]\). Thus, this first condition is met.

Test (Condition 2) by calculating the area under the curve via integrating \(f(x) over [1, e^6]\)

fx_area <- integrate(f, lower = 1, upper = e^6)
fx_area$value

#> [1] 1

The area under the curve \(f(x) = \frac {1}{6x}\) is 1. Thus, this second condition is met.

Therefore, the function \(f(x) = \frac {1}{6x}\) is a PDF because it meets the two conditions for a PDF.