Question 01

Use integration by substitution to solve the integral below.

\(\int4e^{-7x}dx\)

Solution:

Let \(u=-7x\) \(\implies\) \(dx=-\frac{1}{7}dz\)

\(4\int e^z \frac{-1}{7}dz=\frac{-4}{7}\int e^zdz=-\frac{4}{7}e^z+C=-\frac{4}{7}e^{-7x}+C\)

Question 02

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt}=-\frac{3150}{t^4}-220\) bacteria per cubic centimeter per day, where \(t\) is the number of days since treatment began. Find a function \(t\) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

Solution:

\(\int\frac{dN}{dt}=\int(\frac{-3150}{t^4} -220)dt=-\frac{12600}{t^3}-220t+C\)

Since \(N(1)=6530\) this can be used to find \(C\).

\(N(1)=6530=-\frac{12600}{t^3}-220(1)+C\)

\(C=6290\)

\(N(t)=-\frac{12600}{t^3}-220t+6290\)

Question 03

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

Solution:

\(\int_{4.5}^{8.5} 2x-9 dx\)

Using integrate function in R

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

Question 04

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

\(y = x2- 2x- 2\)

\(y = x + 2\)

Solution:

\(x+2=x^2-2x-2\\ \implies x^2-3x-4=0\\ \implies(x-4)(x+1)=0\\ \implies x=4, -1\)

So the intersections occur at \(x=-1\) and \(x=4\)

\(Area=\int_{-1}^{4} x+2 dx - \int_{-1}^{4} x^2-2x-2 dx\)

Using integrate function in R

func<-function(x){x+2}
int<-integrate(func, lower=-1, upper=4)
func1<-function(x){x^2 -2*x -2}
int1<-integrate(func1, lower=-1, upper=4)
print(int$value-int1$value)
## [1] 20.83333

plot the area

plot (func, -5, 7)
plot (func1, -2, 5, add=TRUE)

Question 05

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:

Let C be cost, n be the number of orders per year and x be the lot size.

\(n*x=110 \implies x=110/n\)

\(C=8.25∗n+3.75∗x/2\)

Assuming half an order is in storage at on average

\(C=8.25*n + 3.75*(110/n)/2\)

\(\implies C=8.25*n + 206.25/n\)

\(dC/dn=8.25-206.25/n^2\)

Setting C=0,

\(8.25-206.25/n^2=0\)

\(n=\sqrt \frac{206.25}{8.25}\)

Therefore, number of orders is 5

\(x=110/5=22\)

Lot size is 22

Question 06

Use integration by parts to solve the integral below.

\(\int ln(9x) x^6 dx\)

Solution:

Let \(u=ln(9x)\) and \(v=\frac{1}{7}x^7\)

\(du/dx=1/x \implies du=\frac{1}{x}dx\)

Using the formula \(\int u dv=uv-\int vdu\), we get

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

\(=\frac{ln(9x)x^7}{7}-\int \frac{x^7}{7x}dx\)

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

\(=\frac{ln(9x)x^7}{7}-\frac{1}{7}(\frac{1}{7}x^7 + C)\)

\(=\frac{ln(9x)x^7}{7}-\frac{1}{49}x^7 +C\)

Question 07

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:

The integral of a PDF must be equal to 1:

\(\int_{1}^{e^6} \frac{1}{6x}dx\)

\(=\frac{1}{6} \int_{1}^{e^6} \frac{1}{x}dx\)

\(=\frac{1}{6}ln(x)|_{1}^{e^6}\)

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

\(=\frac{1}{6}(6-0)\)

\(=1\)

f(x) is a probability density function on the interval [1,e6]

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