Question 1:

Use integration by substitution to solve the integral below.

\[ \int4e^{-7x} \ dx \]

Solution:

We will denote u with -7x to apply the integration by substitution.

\[ u = -7x \]

Solving for dx

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

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

Now, we substitute u and dx into the integral:

\[ \int4e^{-7x} \ dx = \int4e^{u}*\frac{du}{-7} \]

\[ \int4e^{-7x} \ dx = \frac {-4}{7}\int e^{u}*{du} \]

\[ \int4e^{-7x} \ dx = \frac {-4}{7}e^{u}+C \]

\[ \int4e^{-7x} \ dx = \frac {-4}{7}e^{-7x}+C \]

where C is the constant of integration

Question 2:

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

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

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.

Solution:

We will integrate the given rate of change of contamination with respect to time t to find find the function N(t) that estimates the level of contamination.

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

First find N(t), where \[ N(1)=6530\]

Next, integrate the given rate function.

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

\[ \int(-3150{t^4-220)} \ dt \]

\[ \frac{-3150}{5}t^5-220t+C\]

Now substitute the initial condition N(1)=6530 to find constant C:

\[ N(1) = \frac{-3150}{5}(1)^5-220(1)+C=6530\]

\[ \frac{-3150}{5}-220+C=6530\]

\[ \frac{-3150}{5}-220+C=6530\]

\[ -630-220+C=6530\]

\[ -850+C=6530\]

\[C=6530+850\]

\[C=7380\]

Now function N(t) which estimates the level of contamination at any given time t is the number of days since treatment began.

\[ N(t) = \frac{-3150}{5}(t)^5-220(t)+7380\]

Question 3:

Find the total area of the red rectangles in the figure below, where the equation of the line is

\[f(x) = 2x-9 \]

Solution:

There are 4 red rectangles that start at 4.5 and end at 8.5

func_3 <- function(x)
{
  2 * x - 9
}
integrate(func_3, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

The total area of the red rectangles is 16

Question 4:

Find the area of the region bounded by the graphs of the given equations:

\[ y=x^2-2x-2, y=x+2 \]

Solution:

First find the points of intersection between the two curves to determine the limits of integration.

\[ x^2-2x-2 = x+2 \]

\[ x^2-3x - 4 = 0 \]

Find the factors of quadratic equations

\[ (x-4)(x+1)=0 \]

Solutions are:

\[ x=4 \]

\[ x=-1 \]

Now integrate the difference of the upper and lower curve

\[ y=x +2 \]

\[ y=x^2 +2x-2 \]

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

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

\[ A = \int_{-1}^{4}(x^2+3x+4)dx \]

\[ A = [\frac{-x^3}{3} + \frac{3x^2}{2}+4x]_{-1}^{4} \]

Now replace the upper and lower limit

\[ A = [\frac{-4^3}{3} + \frac{3(4)^2}{2}+4(4)]-[\frac{-(-1)^3}{3} + \frac{3(-1)^2}{2}+4(-1)] \]

\[ A = [\frac{-64}{3} + 24+16]-[\frac{1}{3} + 1.5-4] \]

\[ A = [\frac{-64}{3} + 24+16]+[\frac{-1}{3} - 1.5+4] \]

\[ A = [\frac{-64}{3} + 40-\frac{1}{3} - 1.5+4] \]

\[ A = [\frac{-65}{3} + 42.5] \]

\[ A = -21.6667+42.5 \]

\[ A = -21.6667+42.5 \]

\[ A = 20.8333 \]

The area between the curves from x=−1 to x=4 is 20.8333 square units. We confirm using pracma R module.

library(pracma)
## Warning: package 'pracma' was built under R version 4.3.3
# Define the  function
integrand <- function(x) { (x + 2) - (x^2 - 2*x - 2) }

# Calculate the integral from x = -1 to x = 4
area <- integral(integrand, -1, 4)
area
## [1] 20.83333

Let us plot two functions. f1(x) is in red and f2(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)

Finding roots of the quadratic function f1(x)

roots <- polyroot(c(-2, -2, 1))

Finding the intersection of two functions.

They intersect where f1(x)−f2(x)=0

\[ (x^2 -2x-2)-(x+2)=0x^2-3x-4=0 \]

Finding the roots

intersection <- polyroot(c(-4, -3, 1))

Xs <- c(intersection[1], roots, intersection[2])
Xs
## [1] -1.0000000+0i -0.7320508+0i  2.7320508+0i  4.0000000+0i
Xs <- c(intersection[1], roots, intersection[2])
Xs
## [1] -1.0000000+0i -0.7320508+0i  2.7320508+0i  4.0000000+0i
plot(x1, eq1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, eq2(x1), col="green")
points(Xs, eq1(Xs))
## Warning in xy.coords(x, y): imaginary parts discarded in coercion

## Warning in xy.coords(x, y): imaginary parts discarded in coercion
text(Xs, eq1(Xs), labels = c("a","b","c","d"), pos = 3)
## Warning in xy.coords(x, y, recycle = TRUE, setLab = FALSE): imaginary parts
## discarded in coercion
## Warning in xy.coords(x, y, recycle = TRUE, setLab = FALSE): imaginary parts
## discarded in coercion
abline(h=0)

# Find individual parts
a1 <- integrate(eq2, lower=Xs[1], upper=Xs[4])
## Warning in integrate(eq2, lower = Xs[1], upper = Xs[4]): imaginary parts
## discarded in coercion

## Warning in integrate(eq2, lower = Xs[1], upper = Xs[4]): imaginary parts
## discarded in coercion
a2 <- integrate(eq1, lower=Xs[1], upper=Xs[2])
## Warning in integrate(eq1, lower = Xs[1], upper = Xs[2]): imaginary parts
## discarded in coercion
## Warning in integrate(eq1, lower = Xs[1], upper = Xs[2]): imaginary parts
## discarded in coercion
a3 <- integrate(eq1, lower=Xs[3], upper=Xs[4])
## Warning in integrate(eq1, lower = Xs[3], upper = Xs[4]): imaginary parts
## discarded in coercion
## Warning in integrate(eq1, lower = Xs[3], upper = Xs[4]): imaginary parts
## discarded in coercion
a4 <- integrate(eq1, lower=Xs[2], upper=Xs[3])
## Warning in integrate(eq1, lower = Xs[2], upper = Xs[3]): imaginary parts
## discarded in coercion
## Warning in integrate(eq1, lower = Xs[2], upper = Xs[3]): imaginary parts
## discarded in coercion
# Combining all we will find the following 
area <- a1$value-a2$value-a3$value-a4$value
area
## [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.

Solution:

Number of orders (x) * Lot Size (n) = 110

C = costs

\[ C = 8.25x + 3.75 *\frac {110/x}{2} \]

\[ C = 8.25x + 3.75 *\frac {55}{x} \]

\[ C = 8.25x + \frac {206.25}{x} \]

To minimize costs, set derivative = zero

\[ C' = 8.25 - \frac {206.25}{x^2} \]

\[ 0 = 8.25 - \frac {206.25}{x^2} \]

\[ \frac {206.25}{x^2}=8.25 \]

\[ 206.25 = 8.25x^2 \]

\[ \frac {206.25}{8.25}= x^2 \]

\[ 25= x^2 \]

The number of orders per year that will minimize inventory costs is 5.

The lot size is 22 because 22 x 5 = 110

Question 6:

Use integration by parts to solve the integral below.

\[ A = \int(9x)* x^6dx \]

Solution:

Apply the integration to solve the integral below:

\[ A = \int(9x)* x^6dx \]

Applying the formula

\[ fg' = fg - \int(f'g) \]

\[ f = ln(9x) \]

\[ f' = \frac{1}{x} \]

\[ g' = x^6 \]

\[ g = \frac {x^7}{7} \]

Now substitute the formula

\[ ln(9x)* x^6 = ln(9x)*\frac{x^7}{7}-\int\frac{1}{x}*\frac{x^7}{7}\]

\[ ln(9x)* x^6 = ln(9x)*\frac{x^7}{7}-\int\frac{x^6}{7}\]

\[ ln(9x)* x^6 = \frac {ln(9x)*x^7}{7}-\frac{x^7}{49}\]

\[ ln(9x)* x^6 = \frac {7*ln(9x)*x^7}{49}-\frac{x^7}{49}\]

\[ ln(9x)* x^6 = \frac {x^7(7*ln(9x))-1}{49}+c\]

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.

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

Solution:

\[\int\frac{1}{6x}dx\]

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

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

\[\frac{1}{6}(ln(x)|_{1}^{e^5}\]

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

Now since ln(1)=0, we are left with the following:

\[(\frac{1}{6})(ln(e^6))\]

\[=ln(e^6)=1\]

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

This is the probability density function on the interval[1,e^6]