Data 605 - Assignment 14
Jim Mundy
- Use integration by substitution to solve the integral: \(\int{4e^{-7x}dx}\)
Let \(u=-7x\), then \(du = -7dx\).
\[ \begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} \\ &= \int{\frac{-4}{7}e^u du} \\ &= \frac{-4}{7}e^u+constant \\ &= -\frac{4}{7}e^{-7x}+ constant \end{split} \]
- 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.
\[ \frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220 \\ \int{(\frac{-3150}{t^4}-220) dt} = \frac{1050}{t^3}-220t+C = N(t) \]
Since \(N(1)= 6530\), then
\[ \begin{split} N(t) &= \frac{1050}{t^3}-220t+C \\ N(1) &= 6530 \\ \frac{1050}{1^3}-220\times 1 +C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \end{split} \]
We can estimate contamination with the following: \(N(t) = \frac{1050}{t^3}-220t+5700\)
- Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).
All the square in the graph have an area of \(1\). Each rectangle has a width of \(1\). Counting each vertical rectangle from left to right we calculate the total area as: \(Area=1+3+5+7=16\).
Alternatively, we can use: \(Area = \int_{4.5}^{8.5}{(2x-9)dx} = 16\)
- Find the area of the region bounded by the graphs of the given equations.
\(y_1 = x_1^2 - 2x_1-2\) \(y_2 = x_2 + 2\)
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)
The quadratic function dips below the \(x\) axis so we need four points to evaluate the integrals and find the area - intersections of two functions and roots of the quadratic function.
Roots of the quadratic function \(f_1(x)\):
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.
Four points are as follows…
## [1] -1.0000000+0i -0.7320508+0i 2.7320508-0i 4.0000000-0iplot(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])
# Combine individual parts
# Last integral is negative because it is below the x-axis, so it needs to be negated
area <- a1$value-a2$value-a3$value-a4$value
area## [1] 20.83333\(Area \approx 20.83\)
- 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.
Let \(x\) be a number of flat irons to order.
\(Yearly\ storage\ cost = {Storage\ cost\ per\ iron} \times {Average\ number\ of\ irons\ stored} = 3.75 \times x/2 = 1.875x\)
\(Yearly\ ordering\ cost = {Cost\ of\ each\ order} \times {Number\ of\ orders} = 8.25 \times 110/x = 907.5/x\)
\(Inventory\ cost = Yearly\ storage\ cost + Yearly\ ordering\ cost = 1.875x+907.5/x = f(x)\)
Differentiate and solve at \(0\):
\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \\ 1.875-\frac{907.5}{x^2} &= 0 \\ 1.875&= \frac{907.5}{x^2} \\ 1.875x^2&= 907.5 \\ x^2&= \frac{907.5}{1.875} \\ x&= \sqrt{\frac{907.5}{1.875}} \\ x&=\sqrt{484} \\ x&=22 \end{split} \]
Every order should contain \(22\) irons, so there should be \(110/22=5\) orders.
- Use integration by parts to solve the integral below.
\(\int{ln(9x) \times x^6 dx}\)
Let \(u= ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\).
Let \(\frac{dv}{dx}=x^6\), then \(v = \int{x^6 dx} = \frac{1}{7}x^7\).
Using the formula for integration by parts: \(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)
\[ \begin{split} \int{ln(9x) \times x^6 dx} &= \frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^7 \times \frac{1}{x} dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^6 dx} \\ &=\frac{7}{49}x^7 \times ln(9x) - \frac{1}{49}x^7 + constant \\ &=\frac{1}{49}x^7 (7ln(9x) - 1) + constant \\ \end{split} \]
- Determine whether \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral.
\(f(x) = \frac{1}{6x}\)
\[ \begin{split} \int_1^{e^6}\frac{1}{6x} dx &= \frac{1}{6} ln(x)|_1^{e^6} \\ &= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1) \\ &= \frac{1}{6} \times 6 - \frac{1}{6} \times 0 \\ &= 1 \end{split} \] 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 \(f(x)>0\).
If \(f(x)=0\) outside of the given interval, then this function is a probability density function.