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

\[\int 4e^{7x}dx = 4\int e^{7x} dx\] Let \(u = 7x\) Let \(du = 7dx\) Then we can substitute \(\frac{du}{7}\): \[\int e^u du = \frac{1}{7} \int e^u du\] \[\int e^u du = e^u\] So, the result is \(\frac{e^u}{7}\)

Let’s substitute u back in:
\[\frac{1}{7} * e^{7x}\] So, the result is \(\frac{4}{7} e^{7x}\)

Adding the constant, our final answer is:

\[\frac{4}{7} e^{7x} + C\]

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.

\[N= 3150 \frac{t^(-3)}{-3} -220t + C\] \[6530 = -1050(1)^{-3} -220(1) + C\] \[6530 + 1270 = C\] \[7800 = C\] \[N = \frac{-3150}{3y^3}-220t + 7800\] \[N = \frac{-1050}{t^3}-220t + 7800\]

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

\[y = 2x -9\] For: x = 5, y = 1
x = 6, y = 3
x = 7, y = 5
x = 8, y = 7 \[Area = R_1 + R_2 + R_3 + R_4\] \[=1*1 + 1*3 + 1*5 + 1*7 = 1+3+5+7 = 16\]

4. Find the area of the region bounded by the graphs of the given equations: \(y=x^2-2x-2, y=x+2\)

\[y_1 = y_2\] \[x^2-2x-2 = x+2\] \[x^2-3x-4=0\] \[(x-4)(x+1) = 0\] \[x = -1, 4\] \[dA = (y_1 - y_2)dx\] \[A=\int^4_{-1}[3x-x^2+4]dx\] \[\bigg[\frac{3x^2}{2} - \frac{x^3}{3} + 4x\bigg]^4_{-1}\] \[\frac{45}{2} - \frac{65}{3} + 20 = \frac{125}{6}\]

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.

\[=\sqrt\frac{2*110*8.25)}{3.75} = 22\] 22 orders per year will minimize inventory costs.