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Chapter 16: Problem 13

Find the line integrals along the given path \(C.\) \(\int_{C}(x-y) d x,\) where \(C: x=t, y=2 t+1,\) for \(0 \leq t \leq 3\)

Short Answer

Expert verified
The line integral along the path is \(-\frac{15}{2}\).

Step by step solution

01

Parameterize the Path

The path is given by the equations \(x = t\) and \(y = 2t + 1\) as a function of \(t\). This path parameterization describes how \(x\) and \(y\) change as \(t\) varies from 0 to 3.
02

Determine the Differential Components

For the integral \(\int_C (x-y) \, dx\), the differential component \(dx\) is equal to \(dt\) since \(x = t\). Therefore, the integral with \(x\) and \(y\) becomes \(\int_0^3 (t - (2t + 1)) \, dt\).
03

Simplify the Integrand

Substitute \(x = t\) and \(y = 2t + 1\) into the integrand: \(x - y = t - (2t + 1) = -t - 1\).Thus, the integral becomes \(\int_0^3 (-t - 1) \, dt\).
04

Calculate the Integral

Evaluate the integral \(\int_0^3 (-t - 1) \, dt\):First, integrate \[\int(-t) \, dt = -\frac{t^2}{2},\] and \[\int(-1) \, dt = -t.\]Thus, we find:\[\left[-\frac{t^2}{2} - t\right]_0^3.\]
05

Evaluate the Definite Integral

Apply the limits of integration from 0 to 3:\[\left[-\frac{3^2}{2} - 3\right] - \left[-\frac{0^2}{2} - 0\right] = \left[-\frac{9}{2} - 3\right] - [0] = -\frac{9}{2} - 3 = -\frac{15}{2}.\]

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Key Concepts

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

Path Parameterization
In line integrals, parameterizing the path is an essential step. It involves expressing the variables of the path, like \(x\) and \(y\), in terms of a single parameter, often \(t\).
This parameterization transforms the path into a more manageable form. For example, in our exercise, the path is given by:\(x = t\) and \(y = 2t + 1\).
This means as \(t\) varies from 0 to 3, \(x\) and \(y\) will adapt according to these expressions. This simple yet powerful technique breaks down complex paths into linear equations based on \(t\).
  • Choose a parameter, often \(t\), to simplify integration.
  • Express all path variables in terms of \(t\).
  • Define the limits of \(t\) according to the path's boundaries.
Integral Evaluation
Evaluating an integral involves several crucial steps to ensure it's solved correctly. First, substitute the parameterized variables into the integral expression. For our example, the integral \(\int_{C}(x-y) \, dx\) becomes \(\int_{0}^{3}(t-(2t+1)) \, dt\).
This step is about replacing the path variables with their parameter \(t\), aligning with how the path is parameterized.
  • Substitute the parameterized equations into the integral.
  • Simplify the integrand as needed after substitution.
  • Ensure the integral's differential component corresponds to the chosen parameter.
Differential Components
Differential components in line integrals indicate how to integrate with respect to a chosen parameter. In this exercise, \(dx\) reflects the rate of change of \(x\) with respect to \(t\), which simplifies to \(dt\) when \(x = t\).
The differential component effectively translates the problem from a multi-variable point of view to a single-variable context, making integration feasible.
  • The differential \(dx\) aligns with the parameterization \(x = t\).
  • In this context, it simplifies the task to integrating terms involving \(t\) with respect to \(t\).
  • This step ensures consistency and correctness in the integral evaluation.
Definite Integral Calculation
Calculating the definite integral requires evaluating the expression between specified limits. In the given exercise, the limits for \(t\) are from 0 to 3. By integrating the expression \((-t - 1)\) with respect to \(t\), we first determine the antiderivative, resulting in \(-\frac{t^2}{2} - t\).
Then, apply the limits:
- Evaluate at the upper limit (\(t = 3\)) and the lower limit (\(t = 0\)).
- Calculate \([-\frac{9}{2} - 3] - [0]\) to finalize the definite integral evaluation.
- This process results in \(-\frac{15}{2}\), which is the value of the line integral over the path, reflecting the overall accumulation dictated by the path's function.
  • Find the antiderivative of the integrand.
  • Apply the upper and lower limits to the antiderivative.
  • Subtract these values to find the integral's result.

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