Problem 42 - Essential Calculus

Question

In Exercises \(41-44\), set up and evaluate the integrals for finding the area and moments about the \(x\) - and y-axes for the region bounded by the graphs of the equations. (Assume \(\rho=1\).) $$ y=\frac{1}{x}, y=0,1 \leq x \leq 4 $$

Step-by-Step Solution

Answer

The area A is \(ln(4)\) square units, the moment about the x-axis \(M_x\) is -3 units, the moment about the y-axis \(M_y\) is 3 units, and the coordinates for the centroid are \((\frac{3}{ln(4)}, -\frac{3}{ln(4)})\).

Step 1: Set up the integral for the area

We'll start off by setting up and calculating the integral to find the area A of the region. The area is given by the definite integral \(\int_a^b f(x) dx\), the integral of the function from the lower limit to the upper limit of the range of x. In this case, \(f(x) = \frac{1}{x}\) and the limits are 1 and 4. So, the integral for the area is \(\int_1^4 \frac{1}{x} dx.\)

Step 2: Evaluate the integral for the area

The integral \(\int_1^4 \frac{1}{x} dx\) can be evaluated by calculating the antiderivative of \(\frac{1}{x}\) at the limits of 1 and 4. The antiderivative of \(\frac{1}{x}\) is \(ln|x|\). Therefore, the area A = \(\left[ln(4) - ln(1)\right] = ln(4).\)

Step 3: Set up the integral for the moments

The first moments \(M_x\) and \(M_y\) about the x-axis and y-axis, respectively, can be evaluated using the formulas \(M_x = \int_a^b yf(x) dx\) and \(M_y = \int_a^b xf(x) dx\). For \(M_x\), \(yf(x)\) becomes \(\frac{1}{x} * \frac{1}{x} = \frac{1}{x^2}\), and for \(M_y\), \(xf(x)\) is simply 1. So, the integrals are \(M_x = \int_1^4 \frac{1}{x^2} dx\) and \(M_y = \int_1^4 dx.\)

Step 4: Evaluate the integral for the moments

The integral for \(M_x\), \(\int_1^4 \frac{1}{x^2} dx\), is equal to \(-1|x|_1^4 = -1 (4 - 1) = -3\). Similarly, \(M_y = \int_1^4 dx\) equals \(x|_1^4 = 4 - 1 = 3\).

Step 5: Compute the coordinates for the centroid

The coordinates \((\bar{x},\bar{y})\) for the centroid are given by \(\bar{x} = \frac{M_y}{A}\) and \(\bar{y} = \frac{M_x}{A}\). Substituting the values obtained, we get \(\bar{x} = \frac{3}{ln(4)}\) and \(\bar{y} = - \frac{3}{ln(4)}\).

Key Concepts

Area Under the Curve

The area under a curve is crucial when analyzing the region confined by graphs. It essentially gives us a measure of the space enclosed between a curve and an axis. In our exercise, the curve is defined by the function \(y = \frac{1}{x}\) between \(x = 1\) and \(x = 4\). To find this area, we use the concept of definite integrals.
The formula for calculating the area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by the integral \(\int_a^b f(x) \ dx\).

• For this problem, \(f(x) = \frac{1}{x}\) and the limits are from 1 to 4.
• Hence, our integral becomes \(\int_1^4 \frac{1}{x} \ dx\), where we find the antiderivative of \(\frac{1}{x}\), which is the natural logarithm.

Evaluating this, we compute \(\ln(4) - \ln(1) = \ln(4)\), establishing the area under the given curve between these limits.

Moments About Axes

In mathematics, the concept of moments helps to understand how mass is distributed about an axis, much like physical moments in mechanics. Here, we focus on two types of moments - about the x-axis and y-axis.

• Moment About the x-axis (\(M_x\)): This evaluates how the "mass" is distributed along the x-axis. The integral for this is \(\int_a^b y f(x) \ dx\). In the exercise, \(y f(x) = \frac{1}{x^2}\), leading to the integral \(\int_1^4 \frac{1}{x^2} \ dx\). This evaluates to \(-1\) times the difference of the reciprocals evaluated between 1 and 4, resulting in \(-3\).
• Moment About the y-axis (\(M_y\)): It describes distribution along the y-axis using \(\int_a^b x f(x) \ dx\). Here, \(x f(x) = 1\), making the integral \(\int_1^4 \ dx\), equivalent to finding \(x\) between the limits, giving a value of 3.

Moments are highly useful in calculating centroids and understanding balance within a system.

Antiderivative

The antiderivative is the reverse process of differentiation, often used to find the original function from its derivative. It's key for solving integrals. For our function \(y = \frac{1}{x}\), the antiderivative is crucial.
To determine the area under the curve, we need the antiderivative of \(\frac{1}{x}\).

• The integral \(\int \frac{1}{x} \ dx\) results in \(\ln|x| + C\), where \(C\) is the constant of integration.
• In definite integrals, constants cancel out, so we focus on the calculated limit values: \(\ln(4)\) and \(\ln(1) = 0\).

By substituting the limits, we simplify \(\ln(4) - \ln(1)\) to find the definite integral value, necessary for computing areas and moments.

Definite Integral

Definite integrals are fundamental in mathematics, helping to calculate areas, volumes, and other measures under or between curves. Unlike indefinite integrals, which produce a family of functions, definite integrals yield a numerical value.
In our exercise, we used the definite integral \(\int_1^4 \frac{1}{x} \ dx\) to find the area. Here’s what happens:

• Calculate the antiderivative of \(\frac{1}{x}\), which is \(\ln|x|\).
• Evaluate it at the upper limit (4), providing \(\ln(4)\).
• Then at the lower limit (1), which gives \(\ln(1) = 0\).
• The difference \(\ln(4) - 0\) provides a final value of \(\ln(4)\).

This clearly shows how definite integrals supply a finite result representing total area, vital in fields like physics, engineering, and economics.