Problem 1:

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

Solution:

From Theorem 5.1.2, rule 12 we know that \(\frac{d}{dx}e^{x} = e^{x}\)

and the antiderivative is \(\int e^{x}dx = e^{x} + C\)

Let \(u = -7x\) __(1)

Differentiating, we get:

\(\frac{du}{dx} = -7 \ \to \ dx = \frac{du}{-7}\) __(2)

Rewriting the original equation in the form of u and du from (1) and (2), we get:

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

Again from Theorem 5.1.2, rule 1, we know that: \(c * f(x) dx = c * \int f(x) dx\). Using this, we can rewrite above equation as:

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

From Theorem 5.1.2, rule 12 we know that \(\int e^{x}dx = e^{x} + C\). Hence, \(\int e^{u}du = e^{u} + C\)

Hence we get \(\frac{-4}{7}e^{u} + C\) which can be written as \(\frac{-4}{7}e^{-7x} + C\).

Hence, \(\frac{-4}{7}e^{-7x} + C\) is the integral of \(\int 4e^{-7x}dx\).

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 1 day was 6530 bacteria per cubic centimeter.

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

Solution:

To find \(N\), we can take the antiderivative, i.e. the integral.

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

From Theorem 5.1.2, rule 2 we know that \(\int (f(x) - g(x)) dx = \int f(x) dx - \int g(x) dx\)

Hence, \(N = \int (-\frac{3150}{t^4}-220)\text{d}t\) can be expressed as \(\int -3150(t^{-4}) \text{d}t - \int 220\text{d}t\)

From Theorem 5.1.2, rule 5 (the power rule for integration), we get: \(N = \frac{-3150}{-3}(t^{-3}) - 220t + C\).

or \(N = 1050(t^{-3}) - 220t + C\).

Substituting \(N(1) = 6530\) in the integral above we get:

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

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

Hence, the function N(t) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter is given by \(N(t) = -1050(t^{-3}) - 220(t) + 5700\).

Problem 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:

The equation of the line is given as \(2x - 9\), and the boundaries appear to be 4.5 and 8.5. Since we’re looking for the area, we can integrate this function over the identified boundaries.

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

From Theorem 5.1.2, rule 2 we know that \(\int (f(x) - g(x)) dx = \int f(x) dx - \int g(x) dx\) and we also know the power rule.

Applying both, we get:

\[ Area = [(8.5)^2 - 9(8.5) + C] - [(4.5)^2 - 9(4.5) + C] \]

or

\[ Area = [72.25 - 76.5] - [20.25 - 40.5] = 16 \]

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:

We need to plot the two graphs first which can be done as below:

curve(x^2 -2*x-2, lwd = 2, xlim=c(-5, 5))
curve(x+2, lwd = 2, xlim=c(-5, 5), add = TRUE)

To find the intersection point of the two graphs, we equate them

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

which can be written as

\[ x^2 -3x -4 = 0 \]

Factorizing, we get (x-4) * (x-1) = 0. Hence the curves intersect at x = 4 and also at x = -1. Solving for y by substituting in the original set of equations, we get y = 6 and y = 1. Hence the two sets of points are (4, 6) and (-1, 1). Using the boundary values of x to find the area bounded by the curve, we have

\[Area = \int_{-1}^4 [x^2 - 3x - 4 ]dx\] \[\frac{x^3}{3} - \frac{3x^2}{2} - 4x\].

Substituting for the values of 4 and -1 in above, we get area of the curve to be ~20.83.

i.e., \[(\frac{4^3}{3} - \frac{3*4^2}{2} - 4*4) - (\frac{-1^3}{3} - \frac{3*-1^2}{2} - 4*-1)\].

\[(\frac{64}{3} - \frac{48}{2} - 16) - (\frac{-1}{3} - \frac{3}{2} + 4)\].

\[(\frac{128 - 144 - 96}{3}) - (\frac{-2 -9+24}{6})\].

\[(\frac{-112}{6}) - (\frac{13}{6}) = (\frac{125}{6}) = 20.83\].

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:

Let the number of irons per order = n.

The number of orders produced per year = n * s >= 110

Assuming that n/2 flat irons are in storage at any given point, the storage cost for these irons = 3.75 * (n/2)

Order cost = 8.25 * (110/n) and total inventory cost = 3.75 * (n/2) + 8.25 * (110/n)

To minimize, we take the derivative

\(f(n) = 3.75*\frac{n}{2} + 8.25*\frac{110}{n}\)

\(f(n) = 1.875*n + \frac{907.5}{n}\)

\(f'(n) = 1.875 + \frac{907.5}{n^2}\)

To minimize, we set the above equation to 0 which gives us:

\(907.5 = 1.875(n^2)\)

or n^2 = 484. Hence the number or irons per order to minimize costs = 22

Hence the number of orders per year = 110/22 = 5

Problem 6:

Use integration by parts to solve the integral below.

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

Solution:

Integration by parts can be expressed as: \[ \int u \ dv = uv - \int v \ du \]

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

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

Plugging this into the integration by parts formula, we get:

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

Separating out the constant, we get:

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

Using the power rule for integration:

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

Hence, the solution is

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

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:

There are two conditions for a probability density function:

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

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