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

\[\int_{}^{}4e^{-7x}dx\] \[ u = -7x \]

\[ \frac {du}{dx} = -7 \]

\[ du = -7dx \]

\[ dx = \frac {du}{-7}\] \[ \frac {-4}{7} \int e^udu \] \[ = \frac {-4}{7} * e^u \] Substitute -7x for u \[ =\frac {-4e^{-7x}}{7} + c\]

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.

\[\frac{dN}{dt} = -\frac{3150}{t^4}-220\] \[\frac{dN}{dt} = -3150t^{-4}-220\]

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

\[ N(t) = \frac {1050}{t^3} - 220t + c\] Given level of contamination after 1 day was 6530 bacteria per cubic centimeter:

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

\[ 6530 = 830 + c \] \[ c = 5700 \]

Thus:

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

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

Based on the above figure, the rectangles start at x = 4.5 and end at x = 8.5

We will integrate the function at these points to find the area.

\[\ \int_{4.5}^{8.5} 2x - 9 dx \] \[=\left. \ {x^2} -9x \right|_{4.5}^{8.5}\]

\[Area = ({8.5^2} -9(8.5)) - ({4.5^2} -9(4.5)) \] \[Area = (72.25- 76.5) - (20.25-40.5)\]

\[Area = (-4.25) - (-20.25)\]

Area of the red rectangles is 16.

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

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

Similar to Discussion 14, we need to find the intersection points first.

\[ x^2 - 2x -2 = x + 2\] \[ x^2 - 3x -4 = 0\]

\[(x-4)(x+1)=0\] x = 4, x = -1

The graph below illustrates these intersection points

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

\[\ \int_{-1}^{4} -x^2+3x+4dx \] \[=\left. \ \frac{-x^3}{3}+ \frac{3x^2}{2} +4x \right|_{-1}^{4}\] \[Area = (\frac{-4^3}{3}+ \frac{3*(4)^2}{2} +4(4)) - (\frac{-(-1)^3}{3}+ \frac{3*(-1)^2}{2} +4(-1)) \] \[Area = (\frac {-64}{3} + \frac{48}{2} + 16) - (\frac{1}{3}+ \frac {3}{2} - 4) \] \[Area = (\frac {-128}{6} + \frac{144}{6} + \frac{96}{6}) - (\frac{2}{6}+ \frac {9}{6} - \frac{24}{6}) \] \[Area = (\frac{112}{6}) - (\frac{-13}{6})\]

Area of the region bounded by the graphs is \(\frac{125}{6}\) or \(20.\overline{83}\)

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.

Number of orders (x) * Lot Size (n) = 110

C = costs

\[ C = 8.25x + 3.75 * \frac {110/x}{2}\] \[ C = 8.25x + 3.75 * \frac {55}{x}\]

\[ C = 8.25x + \frac{206.25}{x}\]

In order to minimize costs, set derivative = zero

\[ C' = 8.25 - \frac{206.25}{x^2}\] \[0 =8.25 - \frac{206.25}{x^2}\]

\[ \frac{206.25}{x^2} = 8.25\]

\[206.25 = 8.25x^2\]

\[\frac{206.25}{8.25} = x^2\]

\[25 = x^2\]

Thus, the number of orders per year that will minimize inventory costs is 5. The lot size is 22 because 22 x 5 = 110

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

\[\int_{}^{}ln(9x)*x^6dx\] Formula \[\int fg' = fg - \int f'g\] \[ f = ln(9x)\] \[ f' = \frac{1}{x}\] \[g' = x^6\] \[g = \frac{x^7}{7}\] Substitute into formula \[\int ln(9x)*x^6 = ln(9x)*\frac{x^7}{7} - \int \frac{1}{x} * \frac{x^7}{7}\] \[ = ln(9x)*\frac{x^7}{7} - \int \frac{x^6}{7}\]

\[ = \frac { ln(9x)* x^7}{7} - \frac {x^7}{49}\] \[ = \frac { 7* ln(9x)* x^7}{49} - \frac {x^7}{49}\]

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

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

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

A function is a probability density function if it integrates over the domain of the variable to 1

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

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

Since \(ln(1) = 0\) , we are left with the following:

\[=(\frac{1}{6}) 6(ln(e)) \]

\[ = ln(e) = 1\]

\(f ( x ) = \frac {1}{6x}\) is a probability density function on the interval \([1, e^6]\)