HW13

Question 1:

Use integration by substitution to solve the integral below.

\[\int 4e^{-7x} dx\]

Let \[u = -7x\] then \[\frac{du}{dx} = -7\]

\[dx = -\frac{du}{7}\]

Substituting these into the integral, we get:

\[\int 4e^u \cdot \left(-\frac{du}{7}\right)\]

\[= -\frac{4}{7} \int e^u du\]

\[= -\frac{4}{7} e^u + C\]

Substituting \[u = -7x\]

\[\int 4e^{-7x} dx = -\frac{4}{7} e^{-7x} + C\]

Therefore, the solution to the integral is:

\[\int 4e^{-7x} dx = -\frac{4}{7} e^{-7x} + C\]

Question 2:

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

\[\frac{dN}{dt} = \frac{-3150t^4}{t^4} - 220\]

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:

\[\frac{dN}{dt}dt= (\frac{-3150t^4}{t^4} - 220)dt\]

\[{dN}= (\frac{-3150t^4}{t^4} - 220)dt\]

\[\int{dN}{dt}= \int(\frac{-3150t^4}{t^4} - 220)dt\] \[{N(t)}=\frac{-3150}{-3t^3} - 220t\]

\[{N(t)}=\frac{1050}{t^3} - 220t+c\]

After 1 day was 6530 bacteria per cubic centimeter:

\[6530=\frac{1050}{1^3} - 220(1)+c\] Where c = 6530-1050+220 = 5700

Result:

\[6530=\frac{1050}{1^3} - 220(1)+5700\]

Question 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 <- function(x) {2*x-9}
f_area <- integrate(f,4.5,8.5)
f_area$value

## [1] 16

Question 4:

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

Enter your answer below.

f(x) = x+2 g(x) = x^2-2x-2

f = g at x = -1 and x = 4

\[\int_{1}^{4} (f(x)-g(x)) dx\]

Substituting the value of f(x) and g(x)

\[\int_{1}^{4} (x+2-(x^2-2x-2)) dx\]

125/6

R code:

f <- function(x) {x^2 -2*x - 2}
g <- function(x) {x + 2}

#retrieve the points where the functions intersect below zero
rt1 <- uniroot(function(x)  f(x) - g(x)  , c(-5,0), tol=1e-8) 
rt1$root

## [1] -1

#retrieve the points where the functions intersect above zero
rt2 <- uniroot(function(x)  f(x) - g(x)  , c(0,5), tol=1e-8) 
rt2$root

## [1] 4

#calculate the area under the f(x)
fx_area <- integrate(f, lower = -1, upper = 4)
fx_area

## -3.333333 with absolute error < 1.2e-13

#calculate the area under the g(x)
gx_area <- integrate(g, lower = -1, upper = 4)
gx_area

## 17.5 with absolute error < 1.9e-13

#Find the difference 
(gx_area$value - fx_area$value)

## [1] 20.83333

Question 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.

Economic Order Quantity

\[EOQ = \sqrt{\frac{2DS}{H}}\]

\[EOQ = \sqrt{\frac{2(110)(8.25)}{3.75}}\]

Number of orders per year to minimize inventory

110/22 

5

R code

a = 3.75/2
b = 8.25*110

(x = sqrt(b/a))

## [1] 22

Number of orders per year to minimize inventory

(110/x)

## [1] 5

Question 6:

Use integration by parts to solve the integral below.

\[\int ln( 9x ) .x^{6} dx\]

u = in9(x)

du = 1/x dx

v = 1/7x^7

dv = x^6dx

Substitute: \[\int ln( 9x ) .x^{6} dx\] \[In9(x).1/7x^7 - \int 1/7x^7 .1/x dx \]

\[1/7x^7.In9(x) - \int x^7/7dx \]

\[1/7x^7.In9(x) - x^7/49 +c\]

Question 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

Answer:

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

\[\int_{1}^{e^6}\frac{1}{6x}\]

\[\frac {1}{6} \int_{1}^{e^6}\frac{1}{x}dx\]

\[\frac {1}{6}.In(x)+c\]

\[\frac {1}{6}.In(e^6)- \frac {1}{6}.In(1)\]

\[\frac {6}{6}- \frac {0}{6}\] 1