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

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

  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.

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

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

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

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

  2. Find the area of the region bounded by the graphs of the given equations. y = x^2-2x-2, y = x + 2 \[ x^{2}-2x-2=x+2 \]

\[ x^{2}-3x-4=0 ; x=4, x=-1\]

\[\int_{-1}^{4} (x+2)-(x^{2} -2x-2)dx \]

\[\int_{-1}^{4} -x^{2} +3x+4 dx \] \[ \frac{-x^3}{3}+3x^{2} +4x \Biggr|_{-1}^{4} =\frac{64}{3}+ \frac{48}{2}+16-(\frac{1}{3}+\frac{3}{2}-4) =\frac{112}{6}-(\frac{-13}{6})=\frac{125}{6} =20.8333 \]

  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.

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

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

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

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

  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. \[ f(x)=\frac{1}{6x} \] To check if a function is a probability density function, it must satisfy: \[1. f(x)>=0 \] \[2. \int_{-\infty}^{\infty} f(x)dx=1 \] Solution:

\[ \int_{1}^{e^{6}} \frac{1}{6x}dx=\frac{1}{6} \int_{1}^{e^{6}} \frac{1}{x} dx=\frac{1}{6}ln(x) \Biggr|_{1}^{e^{6}} =1\]