Question 1

Use integration by substitution to solve the integral below

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

Let u = -7x then

du/dx=-7

Integration equation will become

\[ \int_{}^{}4e^{u}\dfrac{du}{-7}\] \[ = \dfrac{-4}{7}*\int_{}^{}4e^{u}du\] \[ = \dfrac{-4}{7}*e^u + c\] Substitute u = -7x

\[ = \dfrac{-4}{7}*e^-7x + c\]

The Answer is \[\dfrac{-4e^-7x}{7} + c\]

Question 2

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

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

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.

Lets start with simplyfying the equation

\[\dfrac{dN}{dt}=-3150t^-4-220\] Integration of above equation will return the function N(t)

\[N(t)=\int_{}^{}-3150t^-4-220 dt\] \[N(t)=\dfrac{-3150t^-3}{-3}-220t +c\] \[N(t)={1050t^-3}-220t +c\] To find the constant c of the equation we need to use the equation

From the question N(t)=6530, t=1

\[6530 = 1050(1)^-3-220(1)+c\] \[6530 = 1050-220+c\] \[c = 6530 - 830\] \[c = 5700\]

Apply c in the N(t) function to get N(t) for the given problem

\[N(t)={1050t^-3}-220t +5700\]

Question 3

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

The red rectangles lower limit is 4.5 The red rectangles upper limit is 8.5

Integration of the function with the above limits would provide the answer.

gn_funct <- function(x) {2*x-9}
area_fn <- integrate(gn_funct,4.5,8.5)
area_fn
## 16 with absolute error < 1.8e-13

The answer is 16 with absolute error < 1.8e-13

Question 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.

Given two equations

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

Solving the above equation will provide us the intercept points of the function

x = -2, x = 5

By using the above two points we can plot the graph in R

limits <- seq(-2,5,0.5)

y1 <- function(x) {x^2-2*x-2}
y2 <- function(x) {x+2}
plot(limits, y1(limits), type='l', col="green", xlab="x-Axis", ylab="y-Axis")
lines(limits, y2(limits), type='l', col="red", xlab="x-Axis", ylab="y-Axis")
abline(h=0)

To find the areas we need to find the intersecting points

The points are -1, 4

Integrating the function with the above limits would provide the areas

gn_funct <- function(x) {x^2-2*x-2}
area_fn <- integrate(gn_funct,-1,4)
area_fn
## -3.333333 with absolute error < 1.2e-13

The area for the plot \[y=x^2-2x-2\] is

-3.333333 with absolute error < 1.2e-13

gn_funct <- function(x) {x+2}
area_fn <- integrate(gn_funct,-1,4)
area_fn
## 17.5 with absolute error < 1.9e-13

The area for the plot \[y=x+2\] is

17.5 with absolute error < 1.9e-13

Total Area = 17.5+3.3 = ~20

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.

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

Simplyfying above would give us

\[C = 8.25*x+\dfrac{206.25}{x}\]

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

Let Cā€™=0

\[8.25-\dfrac{206.25}{x^2}=0\] \[8.25x^2-206.25=0\] \[x^2=206.25/8.25\] \[x^2=25\] \[x=5\]

Number of orders to minimize inventory costs is 5 and lot size is 110/5 = 22.

Question 6

Use integration by parts to solve the integral below.

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

The Integration by parts formula is

\[ \int_{}^{}udv\] = uv - \[ \int_{}^{}vdu\]

\[ u = ln(9x) \] \[ dv = x^6dx \] \[ du/dx = 1/x \] \[ v = \int_{}^{}x^6dx \]

Substituting above in the formula would give us

\[ \int_{}^{}ln(9x)x^6dx = ln(9x)\int_{}^{}x^6dx - \int_{}^{}\int_{}^{}x^6dxdx/x\]

Solving the above equation would provide us the final answer

\[\dfrac{x^7}{7}(ln(9x)-\dfrac{1}{7})\]

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.

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

Lets test this by integrating the function with the limits

gn_funct <- function(x) {1/(6*x)}
area_fn <- integrate(gn_funct,1,exp(6))
area_fn
## 1 with absolute error < 9.3e-05

The value is 1 which shows the function is PDF within the limits.