1.

Use integration by substitution to solve the integral \(\int 4e^{-7x}dx\)

1. Solution

Let \(u = -7x\). and \(du = -7 dx \ \to \ dx = \frac{du}{-7}\).

Our integral is now \(\int \frac{4e^{u}du}{-7}\). Taking out the constants: \(\frac{4}{-7}\int e^u du\).

Replacing \(u\) leads to: \(\frac{-4}{7}e^{-7x} + C\).

2.

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220\) 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 day 1 was 6530 per cubic centimeter.

2. Solution

the solution will be based on the intergral of this function and can be expressed as: \[ \frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220 \ \to \ \text{d}N = (-\frac{3150}{t^4}-220)\text{d}t \]

\(N = \int (-\frac{3150}{t^4}-220)\text{d}t \ \ = \int -3150(t^{-4}) \text{d}t - \int 220\text{d}t\)

then: \(N = \frac{-3150}{-3}(t^{-3}) - 220t + C\).

Solving for \(N(1) = 6530\):

\(N(1) \frac{-3150}{-3}(1^{-3}) - 220(1) + C = 6530\)

\(C = 6530 - 1050 + 220 = 5700\).

\(N(t) = -1050(t^{-3}) - 200(t) + 5700\).

3.

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

3. Solution

To find the total area we can do the following integration:

\[ \int_{4.5}^{8.5}(2x - 9)dx \]

with power rule:

\[ (x^2 - 9x)\Big|_{4.5}^{8.5} = \Big[(8.5)^2 - 9(8.5)\Big] - \Big[(4.5)^2 - 9(4.5)\Big] \\ = [72.25 - 76.5] - [20.25 - 40.5] = 16 \]

4.

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

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

4. Solution

Weโ€™re looking for the area between the two functions. The intersection points are (-1, 1) and (4, 6), which will serve as the boundaries.

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

\(\approx\) 20.8333333

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.

5. Solution

\(TC = PD + \frac{DK}{Q} + \frac{hQ}{2}\).

\(\frac{\text{d}}{\text{d}Q} = -\frac{DK}{Q^2} + \frac{h}{2}\).

setting to 0 yields:

\(-\frac{DK}{Q^2} + \frac{h}{2} = 0\).

\(Q^{*2} = \frac{2DK}{h} \ \to \ Q^* = \sqrt{\frac{2DK}{h}}\)

plug and play:

\(D = 110\).

\(K = 8.25\).

\(h = 3.75\).

\[ Q^* = \sqrt{\frac{2\cdot 110\cdot 8.25}{3.75}} = \sqrt{\frac{1815}{3.75}} = \sqrt{484} = 22. \]

We found 22 to be the lot size per order.

We are given that the store expects to sell \(n = 110\) flat irons. If there are \(x\) number of irons in each order, our equation is \(22\cdot x = 110 \ \to \ x = \frac{110}{22} = 5\).

5 orders of 22 irons are needed per year

6.

Use integration by parts to solve the integral below: \[ \int \ln(9x)\cdot x^6 \ dx \]

6. Solution

The formula for integration by parts is: \[ \int u \ dv = uv - \int v \ du \]

Let \(u = \ln(9x)\). Using the chain rule \(du = \frac{1}{9x} \cdot 9 \ dx = \frac{1}{x} \ dx\).

Let \(dv = x^6\), then \(v = \int x^6 = \frac{x^7}{7}\).

Plug and Play:

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

removing constant:

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

Power Rule:

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

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

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.

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

7. Solution

There are two conditions for a probability density function:

  • \(f(x) \geq 0 \ \ \ \forall x\)
  • \(\int_{-\infty}^{\infty}f(x) \ dx = 1\)

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

Since the area sums up to 1, we conclude that \(f(x)\) is a pdf on the interval \([1, e^6]\).