91Ó°ÊÓ

Parametrization is a powerful technique in calculus used to simplify complex integrations, especially those involving curves. When dealing with line integrals, we often encounter curves represented by equations like \( y = x^2 \). This presents a challenge when integrating directly with respect to \(x\) or \(y\). By using a parameter, usually denoted as \(t\), we express both \(x\) and \(y\) in terms of \(t\).

For example, in the exercise, the curve \( y = x^2 \) can be parametrized by setting \( x = t \) and \( y = t^2 \). This transformation allows us to work within a familiar interval for \( t \), here from \(-1\) to \(2\). This process not only simplifies the expression but maintains the geometric nature of the problem.

• Convert any complex equation into simpler forms using parameters.
• Set limits for the parameter, which directly translate to limits in integration.
• Ensure that all expressions are consistently transformed, including differential elements.

This is a crucial step because a well-chosen parametrization can make integration straightforward, reducing the problem to a single-variable calculus scenario.

Curve integration, particularly line integrals, is central when working with vector fields and paths in vector calculus. Understanding curve integration helps evaluate functions over a path, instead of along the usual axis. In our scenario, we aimed to evaluate the line integral of \( \int_{C} x y \, dx + (x+y) \, dy \) along the curve \( y = x^2 \).

To start, after parametrizing the curve as \( x = t \), \( y = t^2 \), we substitute these into the integral. This creates a new expression involving the derivatives \( dx = dt \) and \( dy = 2t \, dt \).

By substituting:

• \( x = t \)
• \( y = t^2 \)
• \( dx = dt \)
• \( dy = 2t \, dt \)

the line integral becomes \( \int_{-1}^{2} (t t^2 \, dt + (t + t^2) \, 2t \, dt) \). Thus, the complicated path \( C \) is simplified, allowing a direct calculation using standard integral techniques.

This step allows us to deal with integrals involving paths not parallel to axes, using the outlines provided by the curve itself in the parameter domain.

The Fundamental Theorem of Calculus plays a pivotal role in simplifying the evaluation of integrals, turning complex definite integrals into a simple subtraction of two function values. Once we have parametrized the curve and substituted all necessary components into the integral, we end with an expression \( \int_{-1}^{2} (3t^3 + 2t^2) \, dt \).

Applying the Fundamental Theorem, we find the antiderivative, \( \frac{3}{4}t^4 + \frac{2}{3}t^3 \). The final step involves evaluating this antiderivative at the endpoints, \(-1\) and \(2\).

Calculate:

• For \( t = 2 \): \( \frac{3}{4}(2)^4 + \frac{2}{3}(2)^3 \)
• For \( t = -1 \): \( \frac{3}{4}(-1)^4 + \frac{2}{3}(-1)^3 \)

Subtract these results to find the solution to the integral: \( \left( \frac{52}{3} \right) - \left( \frac{1}{12} \right) = \frac{207}{12} \).

This theorem thus streamlines the final evaluation process, turning multi-step integration problems into a more manageable exercise, relying entirely on evaluating the differences of the function's antiderivatives.