Homework: 13

1. Use integration by substitution to solve the integral below

\[ A = \int\limits_{}^{}4e^{-7x}\, dx = 4\int\limits_{}^{}e^{-7x}\, dx, \quad Let's \quad set \quad u = -7x\\ \frac{du}{dx}=\frac{d(-7x)}{dx}=-7, \quad dx = \frac{-1}{7}du, \quad A = \frac{-4}{7}\int\limits_{}^{}e^u\, du \\ A = \frac{-4}{7}e^{-7x} \]

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of

dN/dt = (-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

\[ B = \frac{-3150}{t^4} - 220, \quad \int\limits_{}^{}B\, dt = -3150\int\limits_{}^{}\frac{1}{t^4}\, dt - \int\limits_{}^{}220\, dt \\ = -3150\int\limits_{}^{}t^{-4}\, dt - 220t + constant\\ =3150(\frac{1}{3t^3}) - 220t + constant\\ N(t)= \frac{1050}{t^3} - 220t +constant\\ at \quad t = 1, N(1) = 6530 \quad \Longrightarrow 6530 = 1050-220+constant \quad \Longrightarrow constant = 5700\\ N(t) = \frac{1050}{t^3} - 220t + 5700 \]

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

f(x) = 0 , x = 4.5, the x-vlaue at the rigth angle will be approximate to 8.5

f <- function(x) x*2 - 9
area = integrate(f, lower = 4.5, upper = 8.5)
cat("The total area of the red rectangles in the area delimited by f ( x ) = 2x - 9, x = 4.5 and x = 8.5 is: ")
## The total area of the red rectangles in the area delimited by f ( x ) = 2x - 9, x = 4.5 and x = 8.5 is:
area
## 16 with absolute error < 1.8e-13

4. Find the area of the region bounded by the graphs of the given equations y1 = x^2 - 2x - 2, y2 = x + 2

x <- seq(-5,5,0.01)
plot(x, (x^2 - 2*x - 2),
main="y2 - y1",
ylab="y",
type="l",
col="blue")
lines(x, x+2, col="red")
#lines(0, 0, col = "black")
abline(h = -5 , v = c(-1, 4), col = "green")

f <- function(x) (-x^2 +3*x+4)
area = integrate(f, lower = -1, upper = 4)

cat("The total area of the red rectangles in the area delimited by equations y1 = x^2 - 2x - 2, y2 = x + 2 is: ")
## The total area of the red rectangles in the area delimited by equations y1 = x^2 - 2x - 2, y2 = x + 2 is:
area
## 20.83333 with absolute error < 2.3e-13

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.

This problem is a little unrealistic because in order to minimize the inventory cost, we need to the revenue or profit estimate per year , then a fixed rate on the profit margin can then be applied to keep inventory cost to reasonable. Also, on the expectation/estimate of sale , there is some margin (maybe 2 to 10%) to account for loss due to damage or manipulation.

inventory cost = 8.25m + 3.75n , where m = number of order per year and n = number of flat iron per year. lot size s = n/m , n = 110 cost = 8.25(n/s) + 3.75n = 8.25(100/s) + 3.75110…this cost inventory function is based on lot size per order and per year. cost (s) = 825*s^-1 + 412.5…function cost(3) may have a max and min…we can find these values by deriving cost(s) and set cost(s)’ = 0, cost(s)’ = -825s^-2 …huummm something not right here.

6. Use integration by parts to solve the integral below

\[ A = \int\limits_{}^{}ln(9x).x^{6}\, dx , \quad Let's \quad set \quad u = ln(9x)\\ \frac{du}{dx}=\frac{d(ln(9x)}{dx}=\frac{1}{9x}, \quad dv = x^6dx, v = \frac{1}{7}x^7\\ \int\limits_{}^{}u\, dv = uv - \int\limits_{}^{}v\, du \\ A = \int\limits_{}^{}ln(9x).x^{6}\, dx = ln(9x).(\frac{x^7}{7})-\int\limits_{}^{}\frac{x^7}{9*7*x}\, dx\\ = ln(9x).(\frac{x^7}{7})-\int\limits_{}^{}\frac{x^6}{63}\, dx = ln(9x).(\frac{x^7}{7})-\frac{x^7}{7*63} +constant\\ A = \frac{x^7}{7}(ln(9x)-\frac{1}{63}) + constant \]

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)= 1/6x.

A probability density function must satisfy \[P(x \epsilon B)=\int\limits_{B}^{}P(x)\,dx \geq 0 \quad for \quad all \quad x \\ \quad and \quad is \quad constrained \quad by \quad the \quad normalization \quad condition, \quad P(-\infty<x<\infty) = \int\limits_{-\infty}^{\infty}P(x)\,dx = 1.\]

## 
## The value of the definite integral. f(x)= 1/6x
## 1 with absolute error < 9.3e-05
## 
## f ( x ) is a probability density function on the interval [1,e^6]