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When we talk about finding the area under a curve, we are essentially looking to determine the space between a function and the x-axis over a specified interval. In many mathematics problems, particularly when dealing with functions and their graphs, understanding the area under the curve is crucial. This concept is not just about shapes or regions on a graph, but also about the integral of a function over an interval.

In our given problem, we have a function defined as \( f(x) = 27 - 3x^2 \). The task is to find the area from \( x = 1 \) to \( x = 3 \). To achieve this, we set up a definite integral, which serves as the mathematical tool required to calculate this area accurately. By evaluating this integral, we can determine the accumulation of values the function takes over the interval, giving us the precise area under the curve.

An antiderivative is a function whose derivative is the original function you started with. It's essential in integration because finding an antiderivative is a key step in solving an integral. When you integrate a function, you're essentially looking for the accumulation of the function's values, which is represented by the antiderivative.

For instance, consider our function \( f(x) = 27 - 3x^2 \). The antiderivative of this function is calculated using integration techniques like the power rule. This step gives us \( 27x - x^3 + C \), where \( C \) is the constant of integration. However, in definite integrals, we don't need to worry about \( C \) because we're more focused on the difference of values between two points, which effectively cancels out any constant term.

Evaluating a definite integral involves calculating the difference between the values of the antiderivative at the upper and lower limits of the interval. This process allows us to find the exact value of the area we are interested in.

In the problem given, the definite integral is set up as \( \int_{1}^{3} (27 - 3x^2) \, dx \). After finding the antiderivative, \( 27x - x^3 \), the next step is to plug in the limits \( x = 3 \) and \( x = 1 \). We compute \[ \left(27(3) - (3)^3\right) - \left(27(1) - (1)^3\right) \], which simplifies to \( 54 - 26 = 28 \).

This calculation means that the area under the curve from \( x = 1 \) to \( x = 3 \) is precisely 28 square units, reflecting the total change over the interval after integrating.

The power rule for integration is a fundamental tool used to find antiderivatives of power functions easily. It's designed to simplify the integration process by transforming each term in the function according to its exponent. If a function has terms like \( x^n \), the power rule comes into play by guiding us to integrate by increasing the power by one and dividing by the new power.

In our specific problem, we apply the power rule to the function \( 27 - 3x^2 \). For the term \( 3x^2 \), this means increasing the power (from 2 to 3) and dividing by the new power, yielding \(-\frac{x^3}{3}\). This simplifies further calculations and leads to a smooth process of finding the antiderivative as \( 27x - x^3 + C \).

Utilizing the power rule efficiently thus becomes an essential skill in calculus, helping us speed up evaluations and avoiding common pitfalls in integration.