HW13

Calculus with R 1. Use integration by substitution to solve the integral below.

u=−7xdu=−7dx ∫4e−7xdx ∫−47eudu ∫−47eudu −47∫eudu −47eu+C −47e−7x+C

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dNdt=−3150t4−220 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.

dNdt=−3150t4−220

3150∫1t4dt−220∫1dt

3150∫−13t3−220t

-220t+C

N(1)=1050t3−220t+C

6530=105013−220(1)+C

6530=1050−220+C

5700=C

N(t)=1050t3−220t+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 [4.5,8.5].

f <- function(x) 2*x-9
paste0('The area is ',integrate(f,4.5,8.5)$value)

## [1] "The area is 16"

[1] “The area is 16”

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

Bounds: x2−2x−2=x+2 x2−3x−4=0 (x−4)(x+1)=0 x={4,−1}

y_one <- function(x) x+2
y_two <- function(x) x^2 - 2*x -2

int_one <- integrate(y_one,-1,4)
int_two <- integrate(y_two,-1,4)
total_area <- int_one$value - int_two$value

[1] “The area of the region bounded by interval [-1,4] is: 20.833”

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.

assumption: 2 flat irons stored on average x=flat iron orders c = cost function storagecosts=3.75∗x/2=1.875x ordercosts=8.25∗110/x=907.5/x

costfunction=1.875x+907.5/x

c′(x)=1.875−907.5x2

c′(x)=1.875x2=907.5

x2=907.51.875

x=(√907.51.875)

x=22

c=1.875∗22+907.5/22=$82.5

lotsize=110/x=5

6. Use integration by parts to solve the integral below.

∫ln(9x)∗x6dx

d(uv)=udv+vdu

∫d(uv)=uv=∫udv+∫vdu

∫udv=uv−∫vdu

u=ln(9x)

dvdx=x6

du=99xdx=1xdx

dv=x6dx

v=17x7

ln(9x)17x7−∫17x71xdx

ln(9x)x77−x749−C

x7(7ln(9x)−1)49+C

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)=16x

f_sev <- function(x) 1/(6*x)
is_pdf <- integrate(f_sev,1,exp(6))

[1] “Since the integral equals 1 from 1 to e^6 it is a probability density function.”