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The term "definite integral" is a fundamental concept in calculus, serving as a way to calculate the accumulated area under the curve of a function. When you have a function, say \( f(z) = 3z^2 - 2z \), and you're interested in analyzing it over a specific range, like from \( z = -1 \) to \( z = 2 \), the definite integral is the tool to use. This is represented as:

• \( \int_{-1}^{2} (3z^2 - 2z) \, dz \)
• This notation \( \int \) symbolizes integration, the process of finding the total area under the curve \( f(z) \).

The definite integral computes this area by summing up an infinite number of infinitesimally thin rectangles under the curve from \( z = -1 \) to \( z = 2 \).
Once evaluated, this integral gives us a numerical value, which here is 6. This value represents the net area between the curve and the z-axis, accounting for areas both above and below the axis.

An antiderivative, also known as a primitive function, is essentially the inverse of taking a derivative. When dealing with integrals, finding an antiderivative is a critical step. For the function \( 3z^2 - 2z \), you first need to determine its antiderivative to evaluate the integral from \( -1 \) to \( 2 \).

• The antiderivative of \( 3z^2 \) is \( z^3 \).
• The antiderivative of \( -2z \) is \( -z^2 \).
• Thus, the antiderivative of \( 3z^2 - 2z \) becomes \( z^3 - z^2 \).

Finding the antiderivative allows us to apply the Fundamental Theorem of Calculus, which connects derivatives and integrals, letting you evaluate the definite integral from a to b. By calculating \( \left[z^3 - z^2 \right]_{-1}^{2} \), we find the precise area under the original function’s curve within the given interval.

When solving calculus problems such as finding the average value of a function, it's crucial to follow a structured problem-solving approach. Here’s how you can tackle such problems efficiently:

• First, identify the function and the interval over which you'll be working. In this problem, it's \( f(z) = 3z^2 - 2z \) over \([-1, 2]\).
• Set up the definite integral for the function over the specified interval. This involves organizing the integral as \( \int_{-1}^{2} (3z^2 - 2z) \, dz \).
• Find the antiderivative, which is essential to computing the integral's value, by solving for the inverse of the derivative process.
• Compute the definite integral by substituting the interval’s bounds into the antiderivative. Here, solving \( \left[z^3 - z^2\right]_{-1}^{2} \) produces 6.
• Finally, calculate the average value by dividing the integral's result by the interval's length \( b - a \), yielding \( \frac{1}{3} \times 6 = 2 \).

By following these steps methodically, you can simplify complex calculus problems into manageable tasks, ensuring thorough understanding and accurate results.