Question 1:

Use integration by substitution to solve the integral below.

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

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

\[ u = -7x, \,du = -7dx, \, -\frac{1}{7}du = dx\]

\[ = 4\int -\frac{1}{7}e^{u}\,du \]

\[ = 4(-\frac{1}{7} \int e^{u}\,du) \]

\[ = -\frac{4}{7} e^u \]

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

Question 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.

\[ N'(t) = -\frac{3150}{t^4} - 220 \]

\[ \int (-\frac{3150}{t^4} - 220)\,dt\]

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

\[ = -\int 3150t^{-4} - 220t \]

\[ = -(-1050t^{-3}) - 220t \]

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

\[ N(1) = 6530 \]

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

\[ 1050 - 220 + C = 6530 \]

\[ C = 5700 \]

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

Question 3:

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

\[ = f(5) \cdot 1 + f(6) \cdot 1 + f(7) \cdot 1 + f(8) \cdot 1 \]

\[ = (2(5) - 9) + (2(6) - 9) + (2(7) - 9) + (2(8) - 9) \]

\[ = 1 + 3 + 5 + 7 \]

\[ = 16 \]

Question 4:

Find the area of the region bounded by the graphs of the given equations:

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

\[ f(x) = x + 2 \]

\[ g(x) = x^2 - 2x - 2 \]

\[ f = g \text{ at } x = -1 \text{ and } x = 4 \]

\[ \int_{-1}^{4} (f(x) - gx))\,dx\]

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

\[ = \left. -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x \right |_{-1}^{4} \]

\[ = -\frac{1}{3}(4^3) + \frac{3}{2}(4^2) + 4(4) - (-\frac{1}{3}(-1^3) + \frac{3}{2}(-1^2) + 4(-1)) \]

\[ = -\frac{64}{3} + 24 + 16 - (\frac{1}{3} + \frac{3}{2} - 4)\]

\[ = -\frac{128}{6} + \frac{144}{6} + \frac{96}{6} - \frac{2}{6} - \frac{9}{6} + \frac{24}{6} \]

\[ = \frac{125}{6} \]

Question 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.

\[ \text{Economic Order Quantity} = EOQ = \sqrt{\frac{2DS}{H}} \]

\[ \text{where}\, D = \text{annual demand,}\, S = \text{fixed purchase order cost,}\, H = \text{annual per unit storage cost} \]

\[ = \sqrt{\frac{2(110)(8.25)}{3.75}} \]

\[ = 22\, \text{units per order, }\, 5\, \text{orders per year} \]

Question 6:

Use integration by parts to solve the integral below.

\[ \int ln(9x) \cdot x^6\,dx \]

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

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

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

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

Question 7:

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

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

\[ \int_{1}^{e^6} \frac{1}{6x}\,dx \]

\[ \frac{1}{6} \int \frac{1}{x}\,dx = \frac{1}{6} \cdot ln(x) + C \]

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

\[ = \frac{6}{6} - \frac{0}{6} \]

\[ = 1 \]

Yes, \(f(x)\) is a probability density function on the interval \(\left[1, e^6\right]\) because the area underneath the curve is 1.