Exercise 1

Use integration by substitution to solve the integral below.

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

Solution 1

Let u=−7x, then du=−7dx.

\[ \begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} &= \int{\frac{-4}{7}e^u du} &= \frac{-4}{7}e^u+constant &= -\frac{4}{7}e^{-7x}+ constant \end{split} \]

Exercise 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.

Solution 2

\[ \frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220 \\ \int{(\frac{-3150}{t^4}-220) dt} = \frac{1050}{t^3}-220t+C = N(t) \]

\[ \begin{split} N(t) &= \frac{1050}{t^3}-220t+C \\ N(1) &= 6530 \\ \frac{1050}{1^3}-220\times 1 +C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \end{split} \]

The level of contamination can be estimated by the following function:

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

Exercise 3

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

Solution 3

\(Area = \int_{4.5}^{8.5}{(2x-9)dx} = 16\)

Exercise 4

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

\(y_1 = x_1^2 - 2x_1-2\)

\(y_2 = x_2 + 2\)

Solution 4

Plotting two functions:

curve(x^2 -2*x-2, -4, 6)
curve(x+2, add = TRUE, -4,6)

\(Area = \int_{-1}^{4}{(x+2)dx} - \int_{-1}^{4}{(x^2 - 2x - 2)dx}\)

\(Area = \int_{-1}^{4}{[(-x^2+3x+4)]dx}\)

Area = [−x3/3+3x2/2+4x]|4−1

Area = (−4^3 / 3 + 3.4^2 / 2 + 4.4) − ( − (−1)^3 / 3 + 3(−1)^2 / 2 + 4.(−1))

Area = 20.83

Exercise 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

C - cost

n - number of orders per year

x - number of irons in an order

Note, we assume that on average half an order is in storage

y * x = 110

x = 110/n

C = 8.25 * n + 3.75 * x/2,

C = 8.25 * n + 3.75 * (110/n)/2,

C = 8.25 * n + 206.25/n

n = 5

x=110/5=22

5 orders per year with a lot size - 22 will minimize inventory costs.

Exercise 6

Use integration by parts to solve the integral below.

\(\int{ln(9x) \times x^6 dx}\)

Solution 6

Let \(u= ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\).

Let \(\frac{dv}{dx}=x^6\), then \(v = \int{x^6 dx} = \frac{1}{7}x^7\).

Using the formula for integration by parts: \(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)

\[ \begin{split} \int{ln(9x) \times x^6 dx} &= \frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^7 \times \frac{1}{x} dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^6 dx} \\ &=\frac{7}{49}x^7 \times ln(9x) - \frac{1}{49}x^7 + constant \\ &=\frac{1}{49}x^7 (7ln(9x) - 1) + constant \\ \end{split} \]

Exercise 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

\[ \begin{split} \int_1^{e^6}\frac{1}{6x} dx &= \frac{1}{6} ln(x)|_1^{e^6} \\ &= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1) \\ &= \frac{1}{6} \times 6 - \frac{1}{6} \times 0 \\ &= 1 \end{split} \]

In this case the sum of the definite integral for the function f(x) on the interval [0,e^6] is 1. Therefore, we can say that the function f(x) is a probability density function.