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Mass of one-dimensional objects Find the mass of the following thin bars with the given density function. $$\rho(x)=2-\frac{x}{2}, \text { for } 0 \leq x \leq 2$$

Short Answer

Expert verified
Answer: The mass of the one-dimensional object is 3 mass units.

Step by step solution

01

Identify the Problem

We are asked to find the mass of a one-dimensional object using its density function: \(\rho(x) = 2 - \frac{x}{2}\) for \(0 \leq x \leq 2\). We need to calculate the integral of the density function over the interval [0, 2].
02

Write the Integral Formula for Mass

The equation to find the mass of the object is given by: $$M = \int_0^2 \rho(x) \, dx$$
03

Substitute the Density Function

Now substitute the given density function, \(\rho(x) = 2 - \frac{x}{2}\), into the mass equation: $$M = \int_0^2 \left(2 - \frac{x}{2}\right) \, dx$$
04

Integrate the Function

To find the mass, we need to integrate the function over the given interval [0, 2]. Integrate the function, using power rule for integration: $$M = \left[ 2x - \frac{x^2}{4} \right]_0^2$$
05

Calculate the Mass M

Now, substitute the values of the interval (2 and 0) into the evaluated integral expression and find the mass: $$M = \left(2(2) - \frac{(2)^2}{4}\right) - \left(2(0) - \frac{(0)^2}{4}\right)$$ $$M = (4 - 1) - (0) = 3$$
06

Write the Final Answer

The mass of the one-dimensional object is 3 mass units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Function
Understanding the density function is crucial when dealing with mass distribution in objects. In the context of a one-dimensional object, the density function, denoted as \( \rho(x) \), represents how the mass is distributed along the object. More specifically, it gives the mass per unit length at a point \( x \) along the object.

For example, in the given exercise, the density function is \( \rho(x) = 2 - \frac{x}{2}\), with \( x \) ranging from 0 to 2. This function implies that the density decreases linearly as \( x \) increases from 0 to 2. At the start (when \( x = 0\)), the density is at its highest, which is 2 mass units per unit length. At the end (when \( x = 2\)), the density is at its lowest, which is 1 mass unit per unit length.

When working with density functions, it is important to note that the actual mass at any point is infinitesimally small since we're dealing with a continuous distribution. To find the total mass, we therefore integrate the density function over the length of the object.
Definite Integral

Fundamentals of Definite Integrals

The definite integral of a function is a fundamental concept in calculus that quantifies the accumulation of quantities, such as mass in this context. A definite integral over an interval \[a, b\] gives the net area under the curve of the function between \( a \) and \( b \).

In our case, the definite integral will be used to calculate the overall mass of the bar. The exercise presents the integral \( M = \int_0^2 \rho(x) \, dx \) where we are integrating the density function over the interval from 0 to 2. This integral effectively 'sums up' the infinitesimal pieces of mass along the bar's length, giving us the total mass.

Students should notice the limits of integration, which correspond to the length of the object. Here, it is evident that the length extends from \( x = 0 \) to \( x = 2 \). These limits are critical as they define the region over which we are accumulating the mass.
Power Rule Integration

Applying the Power Rule for Integration

The power rule for integration is a technique in calculus used to find the integral of a function in the form \( x^n \) where \( n \) is any real number. To integrate such functions, you add 1 to the exponent of \( x \) and then divide by the new exponent.

In our textbook exercise, to find the mass \( M \) of the bar, we need to integrate the density function \( \rho(x) = 2 - \frac{x}{2}\) using the power rule. This involves rewriting the density function as a sum of terms with powers of \( x \) and then integrating each term separately. For instance, \( x\) raised to the first power will integrate to \( \frac{x^2}{2}\), and a constant integrates to the constant times the variable of integration.

The power rule is an elegant and straightforward method of integration, making it particularly useful for functions with polynomial form, which is frequently the case in physics and engineering problems, including working out the mass of one-dimensional objects like rods or wires with variable density.

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