1. Use integration by substitution to solve the integral below.

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

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

4 is constant now we substitute dx with du (reverse chain rule). \[ \frac{-4}{7} \int { { e }^{ u }du } \\ \frac{-4}{7}e^{u}+c\\ \frac{-4}{7}e^{-7x}+c \]

  1. 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:

We integrate the given rate function get the function for the position / level.

The first term can be re-witten with a negative power and then we can apply the power rule for integration.

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

Applying the given initial conditions.
We have N(1)=6530,hence we want to solve for the constant c:

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

The final function becomes:

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

  1. 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 rectangles within the closed interval [4.5,8.5]
By inspection we calculate the area of each rectangle and then add them up. \(A=l*w\)

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

We can also calculate the area by integrating the given function fro 4.5 to 8.5:

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

integrand <- function(x)
  {
  2*x-9
  }
integrate(integrand, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13
  1. Find the area of the region bounded by the graphs of the given equations.

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

Lets visualize

fx <- function(x){ (x^2)-2*x-2 }
gx <- function(x){x+2}
plot (fx, -15, 15)
plot (gx, -5, 5, add=TRUE)

Lets find the points of intersection of these functions which will be the bounds of integration:

We set f(x)=g(x) and solve for x.

\[ x^{2}-2x-2=x+2\\ x^{2}-3x-4=0\\ (x-4)(x+1)=0\\ x=4\\ x=-1 \] The formula for finding the area enclosed by two curves is as follows:

\[ \int _{ a }^{ b }{ (topFunction-lowerFunction) } dx \]

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

Lets integrate and calculate using R:

func <- function(x){  -x^{2}+3*x+4  }
## integrate the function from 0 to infinity
integrate(func, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13
  1. 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

Lets derive the cost function.
Let x be the number of flat irons per order. The number of orders = 110/x.
\[ Yearly\ storage\ cost=Storage cost per iron×Average number of irons stored=3.75×x/2=1.875x \\ Yearly\ ordering\ cost=Cost of each order×Number of orders=8.25×110/x=907.5/x \\ Inventory\ cost=Yearly storage cost+Yearly ordering cost=1.875x+907.5/x=C(x) \] To find the critical points, we derive C(x) and set C’(x)=0. We then determine which critical point minimizes C(x)

\[ C(x)=1.875x + \frac{907.5}{x}\\ C '(x)=1.875-\frac{907.5}{x^{2}} \] Solve for ciritcal points

\[ 0=1.875-\frac{907.5}{x^{2}} \\ \frac{907.5}{x^{2}}=1.875 \\ 907.5=1.875 x^{2} \\ x^2=\frac{907.5}{1.875} \\ x^2=484 \\ x = \sqrt {484} \\ x = \pm22 \] We take the positive value since the number of flat irons cannot be negative.

x=22 minimizes function C(x) Number of orders = 110/22 = 5

  1. Use integration by parts to solve the integral below

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

Solution

We apply integration by parts. dv is the term that can be integrated easily while u is the term that can be derived easily.
\(ln\) term can be derived much easier than integrated and (\(x^6\)) can be derived or integrated with the same ease but in this case we integrate it since we are differentiating the other term.

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

Integration by parts

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

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

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

Solution

Lets 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,\(e^6\)]