Problem 222 - Calculus Volume 1

Question

For the following exercises, find the work done. Compute the work done for a force \(F=12 / x^{2} \mathrm{~N}\) from \(x=1\) to \(x=2 \mathrm{~m}\).

Step-by-Step Solution

Answer

The work done is 6 Joules.

Step 1: Understand the Problem

The problem asks for the work done by a variable force along a straight line over a given distance. The force is given as a function of position, specifically \( F(x) = \frac{12}{x^2} \) N, and we need to calculate the work done from \( x = 1 \) m to \( x = 2 \) m.

Step 2: Introduction to Work Done Formula

To find the work done by a variable force, use the integral of the force over the distance. The formula for the work done, \( W \), is given by the integral \( W = \int_{x_1}^{x_2} F(x) \, dx \), where \( x_1 \) and \( x_2 \) are the initial and final positions.

Step 3: Set Up the Integral

Substitute the given force \( F(x) = \frac{12}{x^2} \) into the integral formula for work. This results in \( W = \int_{1}^{2} \frac{12}{x^2} \, dx \).

Step 4: Evaluate the Integrand

The integrand is \( \frac{12}{x^2} = 12x^{-2} \). Note this form is useful for applying standard integration rules.

Step 5: Integrate the Function

To integrate \( 12x^{-2} \), use the power rule for integration which states \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), for \( n eq -1 \). Applying this formula: \( \int 12x^{-2} \, dx = -12x^{-1} + C = -\frac{12}{x} + C \).

Step 6: Evaluate the Definite Integral

Substitute the limits \( x=1 \) and \( x=2 \) into the antiderivative. Calculate \( W = \left[ -\frac{12}{x} \right]_{1}^{2} = \left( -\frac{12}{2} \right) - \left( -\frac{12}{1} \right) = -6 + 12 = 6 \).

Step 7: Conclude with the Result

The work done by the force over the interval from \( x=1 \) m to \( x=2 \) m is \( 6 \) Joules.

Key Concepts

Integration

Integration is a fundamental concept in calculus. It is the process of finding the integral of a function, which essentially means finding the area under the curve of that function. Consider integration as the reverse of differentiation, where we find the original function given its derivative.
Integration is used in various fields such as physics, engineering, and economics to solve problems involving rates of change and accumulation. In the context of calculating work, integration helps find the total work done by a variable force.

• When a force changes with position, simply multiplying force by distance doesn't apply.
• Integration allows us to sum up the small pieces of work done over each tiny interval of distance.
• This is important in finding accurate results when the force isn't constant.

Using integration, we find the total work performed as the force varies over a specific interval of distance.

Definite Integral

A definite integral is a type of integration that gives you a numerical value, representing the accumulated quantity of interest. It calculates the net area under the curve from one point to another, which can be visualized on the x-axis as the limits of integration.
The definite integral of a function from point \( x_1 \) to \( x_2 \), denoted as \( \int_{x_1}^{x_2} f(x) \, dx \), calculates the total change in the function over the interval \([x_1, x_2] \).

• Unlike indefinite integrals, definite integrals have specified limits of integration, which give a precise value.
• This value represents the accumulated quantity, such as area or in our case, the work done.
• When evaluating a definite integral, we find the antiderivative first, then substitute the upper and lower bounds to compute the result.

Thus, in the exercise given, the definite integral from \( x = 1 \) to \( x = 2 \) calculates the total work done by applying the force over the specified range.

Power Rule for Integration

The power rule for integration is one of the foundational rules in calculus. It helps in integrating functions that are powers of \( x \), making it an essential tool for solving many types of integral problems.
According to the power rule, for any real number \( n eq -1 \), the integral of \( x^n \) with respect to \( x \) is:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

• \( C \) is the constant of integration that appears in indefinite integrals.
• This rule simplifies the process of integration by providing a straightforward formula for polynomials and expressions that can be rewritten as polynomials.
• It proves useful in our example, where \( F(x) = \frac{12}{x^2} \) is expressed as \( 12x^{-2} \).

Applying the power rule to \( 12x^{-2} \) involves increasing the exponent by one and dividing by the new exponent, resulting in \( -\frac{12}{x} + C \). This application demonstrates how the power rule facilitates finding antiderivatives, which are crucial for calculating definite integrals.