HW13

1) Integration by substitution

Use integration by substitution to solve the integral below.

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

  • u = \({e^{-7x}}\)
  • \({{4}\int e^{-7x}}\)

2) Functions

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \({\frac{dN}{dt}=\frac{3150}{t^4}-2}\) 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.

f(x)= \({{6530}{t^{1/4}}-2}\) =

  • \({\int_1^tf(x)\frac{dN}{dt}}\)

3) Area Functions

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) = 2x -9
  • Solve for y
  • 2x - 9 = 0
  • x = 4.5
  • y = 2x -9
  • A = 1/2(x*y)
integrand <- function(x) {2*x-9} 
A<-integrate(integrand, lower = 4, upper = 9)
A<-A[1]

Total area is approx 20

4) Graph Areas

Find the area of the region bounded by the graphs of the given equations. y = x2-2x-2, y = x + 2

eq <- function(x){x^2-2*x-2}
eq2<-function(x){x+2}
plot(eq(1:5), type='l')
lines(eq2(1:5), type='l')

integrand <- function(x) {(x^2-2*x-2)-(x+2)} 
A<-integrate(integrand, lower = 1, upper = 4)
A<-A[1]

Area of region is approx -13.5

5) Word Problems

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.

  • Expected Unit Sales = 110
  • Storage Costs = 3.75*(110-UnitsSold)
  • Fixed Costs = -x*8.25
x<-110
eq<- function(x){3.75*(110-x)+x*8.25}
plot(eq(1:100), type='l')

integrand <- function(x) {3.75*(110-x)+x*8.25}
A<-integrate(integrand, lower = 0, upper = 1)
A<-as.double(A[1])
A<-A/100

Selling 4.1475 lots of a 100 will minimize cost

6) Integration II

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

  • u = ln(9x) = \({\frac{9}{x}}\)
  • v = \({x^6dx}\) = \({5x^5}\)
  • = -ln(\({\frac{9}{x}}\))-\({5x^5}\)+C

7) Density Function

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

  • \({\int_0^1\frac{-1}{6x^2}dx}\)
eq<-function(x){1/(6*x)}
eq2<- function(x){exp(x)}
plot(eq(1:100), type='l')

plot(eq2(1:100),type='l')

integrand <- function(x) {-1/(6*x^2)} 
A<-integrate(integrand, lower = 1, upper = 10)
A<-A[1]
E<-integrand(1)
O<-integrand(exp(6))

The integrate is -0.15. The results when x = \({e^6}\) is -0.1666667 and the results for when x =1 is -1.024035410^{-6}. The probability is 0.0166718