Problem 1:

Use integration by substitution to solve the integral below:

\(\int 4e^{-7x}dx\)

Solution 1:

Let \(u = -7x\)

\(du = -7 dx\)

\(dx = \frac{du}{-7}\)

\(\frac{4}{-7}\int e^u du\)

\(\frac{-4}{7}e^{-7x} + C\)

 

Problem 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.

Solution 2:

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

Find \(N\)

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

Apply the power rule:

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

 

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

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

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

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

 

Problem 3:

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

Solution 3:

\(A = \int_{4.5}^{8.5}(2x - 9)dx\)

Total Area = 16

 

Problem 4:

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

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

Solution 4:

curve(x^2 - 2*x - 2, -1, 4, col = 'purple')
curve(x + 2, add = TRUE, -1, 4, col = 'red')

The intersect points (boundaries) in the above graph are (-1, 1) and (4, 6). The equation to find the area of the region bounded by the graphs of the given equations is as follows:

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

Using R, we now perform the calculation to find the area.

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

Answer: The area of the region bounded by the graphs of the given equations is approximately 20.83.

 

Problem 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.

Solution 5:

\(Orders(x) * LotSize(n) = 110\)

Let \(C = cost\)

\(C = 8.25x + 3.75 * \frac{110/x}{2}\)

\(C = 8.25x + 3.75 * \frac{55}{x}\)

\(C = 8.25x + \frac{206.25}{x}\)

Minimize costs by setting the derivative to zero.

\(C' = 8.25 - \frac{206.25}{x^2}\)

\(0 = 8.25 - \frac{206.25}{x^2}\)

\(\frac{206.25}{x^2} = 8.25\)

\(206.25 = 8.25x^2\)

\(\frac{206.25}{8.25} = x^2\)

\(25 = x^2\)

Answer: To minimize inventory costs, the store should make 5 orders of 22 irons per year.

 

Problem 6:

Use integration by parts to solve the integral below:

\(\int \ln(9x)\cdot x^6 \ dx\)

Solution 6:

Let \(u = \ln(9x)\).

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

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

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

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

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

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

 

Problem 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.

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

Solution 7:

In order for \(f(x)\) to be a probability function, the area must sum to 1:

\(F(x) = \int_{1}^{e^6} f(x)dx = 1\)

\((x) = \frac{1}{6x}\)

\(F(x) = \int_{1}^{e^6} \frac{1}{6x}dx\)

\(F(x) = \frac{1}{6} \int_{1}^{e^6} \frac{1}{x}dx\)

\(F(x) = \frac{1}{6} ln(x)|_1^{e^6}\)

\(F(x) = \frac{1}{6} [ln(e^6) - ln(1)]\)

\(F(x) = \frac{1}{6} [6-0] = 1\)

 

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