1. Use integration by substitution to solve the integral below

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

\[u=-7x \\ dx = -\frac{1}{7}du \]

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

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt=−3150t4−220 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.

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

The function N(t) is:

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

For N(t) = 6530 where t = 1

\[ \frac{1050}{1^3} -220(1)+C=6530 \\ 830+C=6530 \\ C=5700 \] \[ N(t)= \frac{1050}{t^3}−220t+5700 \]

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is f ( x ) = 2x - 9. \[ Area =\int_{4.5}^{8.5} (2x-9 )\,dx=(\frac{2x^2}{2}-9x)|_{4.5}^{8.5}=(8.5^2-9 \cdot 8.5) - (4.5^2-9 \cdot 4.5)=16\]

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

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

Find the intersection points:

\[x^2 - 2x - 2 = x + 2 \\ x^2 - 3x - 4=0 \\ (x-4)(x+1)=0 \\ x_1=4, x_2=-1\]

Use the intersection to find the area:

\[ Area =\int_{-1}^{4} (x^2 - 2x - 2)\,dx - \int_{-1}^{4} (x+2)\,dx \\ =(\frac{x^3}{3}-\frac{2x^2}{2}-2x)|_{-1}^{4}-(\frac{x^2}{2}+2x)|_{-1}^{4} \\ =[\frac{4^3}{3}-\frac{2 \cdot 4^2}{2}-2 \cdot4 - \frac{(-1)^3}{3}+\frac{2 \cdot (-1)^2}{2}+2 \cdot (-1)] - [\frac{4^2}{2}+2 \cdot4 - \frac{(-1)^2}{2}+2 \cdot (-1)] \\ =20.83\]

  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.

Cost function:

Let x = lot size

Storage cost = 3.75*x / 2

Order cost = 8.25 * 110/x

Cost function = 8.25(110/x) + 3.75(x/2) = 907.5/x + 1.875x

\[ \frac{dC}{dx} = -1*907.5x^{-1-1}+1.875x^{1-1} \\ = -\frac{907.5}{x^2}+1.875 \]

\[\frac{907.5}{x^2}+1.875 = 0 \\ 1.875=\frac{907.5}{x^2} \\ x^2 = \frac{907.5}{1.875} \\ x=22\]

Lot size = 22 and number of orders = 110/22 = 5

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

\[\int ln(9x) \cdot x^6\,dx\]

\[\int f \, dg=f \cdot g - \int g \, df, \\ f=ln(9x)\\ df=\frac{1}{x}dx \\ g=\frac{x^7}{7} \\ dg=x^6dx \]

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

  1. Determine whether f (x) is a probability density function on the interval [1, e^6]. If not, determine the value of the definite integral.

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

if it is a PDF, the sum of the area should be 1

\[\int_{-1}^{e^6}\frac{1}{6x}\,dx=\frac{ln(x)}{6}|_1^{e^6}=\frac{1}{6}[(ln(e^6)-ln(1)]=\frac{1}{6}(6-0)=1 \]

Therefore, this is a PDF.