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

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



2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt} = -\frac{3500}{{t }^{4 }} - 220\) 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.



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



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

\(y = {x}^{2} -2x - 2, y = x + 2\)



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 X be the units per orders Let C be the total cost 110 is the total number of unit 8.25 is the cost per order 3.75 is the cost per unit

Assuming at any time half of the ordered items are in store

Therefore the total cost can be written as

\[C = 8.25*110/X + 3.75*X/2\] \[C = 907.5/X + 1.875*X\] Finding the derivative \[f^{ 1 }\left( C \right) = \frac{-907.5}{{X}^{2}} + 1.875\] To find least cost, set C = 0 \[0 = \frac{-907.5}{{X}^{2}} + 1.875\] X = 22

Therefore, the lot size is 22 and number of order = 110/22 = 5

6. Use integration by parts to solve the integral below

\[\int ln(9x) . {x}^{6}dx\]

Using the formula for integration by parts: \(\int u.dv = uv - \int v. du\)

Assume \(u = ln(9x)\), hence \(du = \frac{1}{x}dx\)
Assume \(dv = x^6dx \\\), hence \(v = \frac{1}{7}x^7 \\\)

Therefore, \[\int udv = uv - \int vdu \\\] \[= ln(9x)*\frac{1}{7}x^7 - \frac{1}{7}\int x^7.\frac{1}{x}dx \\\] \[= \frac{1}{7}*ln(9x)*x^7 - \frac{1}{7}*\int x^6.dx \\\] \[= \frac{1}{7}x^7 [ln(9x) - \frac{1}{7}]\]

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.

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

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

Hence we can determine that the given function is a probability density function