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

\[ \int 4e^{-7x}dx \\ U = -7x \\ dU = -7dx \\ dx = \frac{dU}{-7} \\ 4\int e^U\frac{dU}{-7} \\ \frac{4}{-7}\int e^UdU \\ \frac{4}{-7} e^U + C \\ \frac{4}{-7} e^{-7x} + C \]

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.

The constant of integration, \(N_0\) is found by setting N(1) = 6530. Note the function is discontinuous at t = 0.

\[ \frac{dN}{dt} =\frac{3150}{t^4}-220 \\ dN = (\frac{3150}{t^4}-220)dt \\ N = \int \frac{3150}{t^4}dt-\int 220dt \\ N = N_0 - \frac{3150}{3t^3} - 220t \\ N = N_0 - \frac{3150}{3t^3} - 220t \\ N(1) = N_0 - \frac{1050}{1^3} - 220(1) \\ N_0 = 6530 + 1050 + 220 \\ N_0 = 7800 \\ N = 7800 - \frac{1050}{t^3} - 220t \]

curve(7800-1050/x^3-220*x, 0.5,35)

It looks as if this model only works on the interval [1,35.45].

3.Find the total area of the red rectangles

in the figure below, where the equation of the line is f(x) = 2x-9.

\[ A_1 = (5.5-4.5)*1 = 1 \\ A_2 = (6.5-5.5)*3 =3 \\ A_3 = (7.5-6.5)*5 = 5\\ A_4 = (8.5-7.5)*7 = 7 \\ A_{total} = 1+3+5+7 = 16 \]

For the exact area:

\[ A = \int_{4.5}^{8.5} 2x-9 dx \\ A = [x^2 - 9x]|_{4.5}^{8.5} \\ A = [8.5^2-9*8.5]-[4.5^2-9*4.5] \\ A = 16 \]

This method gives an excellent estimate for the area under the curve. Non-linear functions may not be as precise.

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

\[ y = x^2 -2x-2 \\ y =x+2 \\ A = \int_{-1}^{4}x+2 dx -\int_{-1}^{4}x^2 -2x-2 dx\\ A = \frac{1}{2}x^2|_{-1}^{4} +2x|_{-1}^{4} -[\frac{1}{3}x^3 - x^2 -2x]|_{-1}^{4} \\ A = -[\frac{1}{3}x^3 - \frac{3}{2}x^2 -4x]|_{-1}^{4} \\ A = [\frac{3}{2}x^2 +4x -\frac{1}{3}x^3]|_{-1}^{4} \]

curve(x^2 -2*x-2, -1, 4)
curve(x+2, add = TRUE, -1,4)

((3/2)*4^2 +4*4 -(1/3)*4^3) - ((3/2)*(-1)^2 +4*(-1) -(1/3)*(-1)^3)
## [1] 20.83333

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.

Let C be cost, r be the number of orders per year, and x be the number of at irons in an order. \(rx = 110\) so \(x = \frac{110}{r}\), assume half an order is in storage at on average. Such that,

\[ C = 8.25r + \frac{3.75x}{2} \\ C = 8.25r+\frac{206.25}{r} \\ C' = 8.25 - \frac{206.25}{r^2} \\ C' = 0 \\ r = \sqrt{\frac{206.25}{8.25}}\\ r = 5 \space\text{orders per year} \]

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

\[ \int ln(9x)*x^6dx \\ U = ln(9x) \\ dU = \frac{1}{x}dx \\ dV = x^6dx \\ V = \frac{1}{7}x^7 \\ \int UdV = UV - \int VdU \\ \frac{1}{7}ln(9x)x^7 - \frac{1}{7}\int x^6dx \\ \frac{1}{7}x^7[ln(9x) - \frac{1}{7}] \]

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.

for f(x) to be a probability density function:

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

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