Question 1

Use integration by substitution to solve the integral below.

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

Solution

Select u to be the g(x) inside f(g(x)) \[ u=-7x\\ du=-7dx\\ -\frac{du}{7}=dx \]

Substitute dx with our du (reverse chain rule) pull out the 4 (constant) from the integrand

\[ \frac{-4}{7} \int { { e }^{ u }du } \\ \frac{-4}{7}e^{u}+c\\ \frac{-4}{7}e^{-7x}+c \]

Question 2

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 N( t ) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

Solution

As rate is provided we need to derive “position” function by evaluting integral of the rate function.

We are re-writing the first term of the integrand using negative power so that we can easily apply the power rule other than the quotient rule for integration. Evaluating the integral term by term and then simplifying

\[ \int {(\frac{-3150}{t^{4}}-220)dt}\\ \int {(-3150{t^{-4}}-220)dt}\\ \int-3150t^{-4}dt - \int -220dt\\ -3150\int t^{-4}dt - 220\int dt\\ -3150 \frac{-1}{3} t^{-3} - 220t\\ N(t)= \frac{1050}{t^{3}}-220t+c \]

After identification N(t) can be calcuated as N(1)=6530,hence we may solve constant c using the initial condition

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

Final answer is

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

Question 3

Find the total area of the red rectangles in the figure below, where the 3. equation of the line is f(x)=2x-9

Solution

We can see 4 rectangles in the given plot of the function within the closed interval [4.5,8.5]

Using fundamental theorem of calucalus or by inspection.

By inspection we may calculate the area of each rectangle and then add them up. A=lxw

\[ A1=(1)(5.5-4.5)=1\\ A2=(1)(6.5-5.5)=3\\ A3=(1)(7.5-5.5)=5\\ A4=(1)(8.5-6.5)=7\\ A=A1+A2+A3+A4=1+3+5+7=16 \]

Using part 2 of the fundamental theorem of calculus

\[ \int _{ 4.5 }^{ 8.5 }{ 2x-9 } dx\\ \]

Use R to evaluate simple integral at the end points

## define the integrated function
integrand <- function(x)
  {
  2*x-9
  }


## integrate the function from 0 to infinity
integrate(integrand, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

Question 4

  1. Find the area of the region bounded by the graphs of the given equations.

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

Soution

Visualize

fx <- function(x)
  {
  (x^2)-2*x-2
}

gx <- function(x)
  {
  x+2
}

plot (fx, -15, 15)
plot (gx, -5, 5, add=TRUE)

Find the area enclosed by f(x) and by g(x). Finding exact bounds of integration.

Set f(x)=g(x) and solve for x . The intersection of two curves are the bounds of integration. We should get two values for x because of the quadratic term.

\[ x^{2}-2x-2=x+2\\ x^{2}-3x-4=0\\ (x-4)(x+1)=0\\ x=4\\ x=-1 \]

The formula for finding the area enclosed by two curves is as follows:

\[ \int _{ a }^{ b }{ (top-bottom) } dx \]

As we know a=-1 and b=4,the top function is the function that encloses the upper half of the area and the bottom function is the function that encloses the bottom of the area. In our case, the top function is the straight line while the bottom function is the parabola

\[ \int _{ -1 }^{ 4 }{ (x+2)-(x^{2}-2x-2) } dx\\ \int _{ -1 }^{ 4 } {(x+2-x^{2}+2x+2)dx}\\ \int _{- 1 }^{ 4 } {(-x^{2}+3x+4)dx} \]

Using R to integrate and calculate

## define the integrated function
integrand <- function(x)
  {
  -x^{2}+3*x+4
  }


## integrate the function from 0 to infinity
integrate(integrand, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13

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

Soutions

This problem is example of simple optimization . In opitmization problems, we need to undertsand the givens and identify relationship between the givens.

We are trying to optimize the cost function but the biggest challenge at the moment is to derive the cost function. We can use the information in the problem statement to help us accomplish this task.

We let x be the number of flat irons per order. The numberof orders is found from the ratio 110/x.

The yearly carrying cost is 3.72 times the number of flat irons, divided by 2. (3.72x)/2

The yearly ordering cost is the number of flat irons expected to be sold multiplied by the fixed cost per order, divded by the number of flat irons (8.25(110))/x

The cost function is assembled from all of these pieces.

We need to derive C(x) and set c’(x)=o, to find the ciritcal points. Once we find the ciritcal points, we determine which one minimizes C(x)

\[ C(x)=1.86x + \frac{907.5}{x}\\ C '(x)=1.86-\frac{907.5}{x^{2}} \] Solve for ciritcal points

\[ 0=1.86-\frac{907.5}{x^{2}}\\ \frac{907.5}{x^{2}}=1.86\\ 907.7=1.86 x^{2}\\ 488.0108=x^{2}\\ +22.09097=x\\ -22.09097=x \] Reject the negative ,hence

\[ +22.09097=x \] We can round this to 22.

Apply second derivative test to check if this is indeed a minimum. We need to evaluate C’’(x)

\[ C ''(x)=\frac{907.5}{x^{3}}\\ C''(22)=\frac{907.5}{22^{3}}>0 \]

Hence we conclude that x=22 minimizes function C(x) 22 flat irons and 5 orders per year minimizes cost

Question 6

Use integration by parts to solve the integral below

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

Soution

Apply following formula for integration by parts. We also need to be strategic when selecting u and dv. We pick dv to be the term that can be integrated easily while u is the term that can be derived easily. For our problem, the ln term can be derived much easier than integrated. The power term can be derived or integrated with the same ease but we should intgrate this term since we ae differentiating the other.

\[ uv-\int vdu \\ u=ln(9x)\\ du=\frac{1}{x}dx\\ dv=x^{6}\\ v=\frac{1}{7}x^{7} \]

By integration by parts, we have the following integral. Simplify and evaluate

\[ \frac{x^{7}ln(9x)}{7}-\int \frac{1}{7}x^{7}\frac{1}{x}dx\\ \frac{x^{7}ln(9x)}{7}-\frac{1}{7} \int \frac{x^{7}}{x}dx\\ \frac{x^{7}ln(9x)}{7}-\frac{1}{7}\int x^{6}dx\\ \frac{x^{7}ln(9x)}{7}-\frac{1}{7}(\frac{x^{7}}{7})+C\\ \frac{x^{7}ln(9x)}{7}-\frac{x^{7}}{49}+C \]

OR

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

Question 7

Determine whether f(x) is a probability density function on the interval 1, e6 . If not, determine the value of the definite integral

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

Soultion

Evaluate the definite integral

\[ \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } dx } \\ \frac{1}{6} \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ x } dx } \\ \frac{1}{6}(ln(e^{6})-ln(1))\\ \frac{1}{6}(6-0)\\ =1 \] Hence, f(x) is a probability density function on the closed interval [1,e6]