Problem 1

Use integration by substitution to solve the integral below.

\(\int{4e^{-7x}dx}\)

Answer to Problem 1

Let \(u=-7x\), then \(du = -7dx\).

\[ \begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} \\ &= \int{\frac{-4}{7}e^u du} \\ &= \frac{-4}{7}e^u+constant \\ &= -\frac{4}{7}e^{-7x}+ constant \end{split} \]

Problem 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.

Answer to Problem 2

\[ \frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220 \\ \int{(\frac{-3150}{t^4}-220) dt} = \frac{1050}{t^3}-220t+C = N(t) \]

Since \(N(1)= 6530\), then

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

It is possible to estimate the level of contamination by the following function:

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

Problem 3

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

Answer to Problem 3

Each square in the graph has an area of \(1\). Each rectangle has a width of \(1\). Counting each rectangle left to right the areas are \(Area=1+3+5+7=16\).

A more elegant and more universal solution can be produced using integral.

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

The little triangles above the line are the same as the missing triangles below the line, so the area of all rectangles is equal to the area under the line from \(4.5\) (where the line intersects the \(x\) axis) to \(8.5\) (the right side of the last rectangle).

Problem 4

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

\(y_1 = x_1^2 - 2x_1-2\) \(y_2 = x_2 + 2\)

Answer to Problem 4

Let us plot two functions. \(f_1(x)\) is in red and \(f_2(x)\) is in green.

eq1 <- function(x) x^2-2*x-2
eq2 <- function(x) x+2

min <- -2
max <- 5
x1 <- seq(min, max, 0.05)
plot(x1, eq1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, eq2(x1), col="green")
abline(h=0)

Finding roots of the quadratic function \(f_1(x)\).

roots <- polyroot(c(-2, -2, 1))

Finding the intersection of two functions. They intersect where \(f_1(x)-f_2(x)=0\).

\((x^2-2x-2)-(x+2)=0\) \(x^2-3x-4=0\)

Find the roots.

intersection <- polyroot(c(-4, -3, 1))

It has four points as provided below

Xs <- c(intersection[1], roots, intersection[2])
Xs
## [1] -1.0000000+0i -0.7320508+0i  2.7320508-0i  4.0000000-0i
plot(x1, eq1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, eq2(x1), col="green")
points(Xs, eq1(Xs))
text(Xs, eq1(Xs), labels = c("a","b","c","d"), pos = 3)
abline(h=0)

The area between two graphs is equal to the area under function 2 from \(a\) to \(d\) minus the area under function 1 from \(a\) to \(b\) and from \(c\) to \(d\) plus the area over function 1 from \(b\) to \(c\).

\(Area = \int_a^d{f_2(x)dx} - \int_a^b{f_1(x)dx} - \int_c^d{f_1(x)dx} + \int_b^c{f_1(x)dx}\)

# Find individual parts
a1 <- integrate(eq2, lower=Xs[1], upper=Xs[4])
a2 <- integrate(eq1, lower=Xs[1], upper=Xs[2])
a3 <- integrate(eq1, lower=Xs[3], upper=Xs[4])
a4 <- integrate(eq1, lower=Xs[2], upper=Xs[3])

# Combining all we will find the following 
area <- a1$value-a2$value-a3$value-a4$value
area
## [1] 20.83333

\(Area \approx 20.83\)

Problem 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.

Answer to Problem 5

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

\(Yearly\ storage\ cost = {Storage\ cost\ per\ iron} \times {Average\ number\ of\ irons\ stored} = 3.75 \times x/2 = 1.875x\)

\(Yearly\ ordering\ cost = {Cost\ of\ each\ order} \times {Number\ of\ orders} = 8.25 \times 110/x = 907.5/x\)

\(Inventory\ cost = Yearly\ storage\ cost + Yearly\ ordering\ cost = 1.875x+907.5/x = f(x)\)

Now we will find the minimized value, differentiate and solve at \(0\):

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

As per the observation there has to be \(22\) flat irons, so there should be \(110/22=5\) orders.

Problem 6

Use integration by parts to solve the integral below.

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

Answer to Problem 6

Let \(u= ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\).

Let \(\frac{dv}{dx}=x^6\), then \(v = \int{x^6 dx} = \frac{1}{7}x^7\).

Formula of integration by parts: \(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)

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

Problem 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.

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

Answer to Problem 7

\[ \begin{split} \int_1^{e^6}\frac{1}{6x} dx &= \frac{1}{6} ln(x)|_1^{e^6} \\ &= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1) \\ &= \frac{1}{6} \times 6 - \frac{1}{6} \times 0 \\ &= 1 \end{split} \] Final Findings : The definite integral of the function on interval \([1, e^6]\) is \(1\). Additionally, if \(x>0\), then \(f(x)>0\), so for this interval \(f(x)>0\). As long as \(f(x)=0\) outside of the given interval, due to this we can can finally agree that it is is a probability density function.