91Ó°ÊÓ

In calculus, a definite integral is a fundamental concept that helps in computing the area under a curve over a specific interval. When you see the notation \( \int_{a}^{b} f(x)\, dx \), it represents the signed area under the curve of the function \( f(x) \) from \( x = a \) to \( x = b \). Here, "signed area" means that areas above the x-axis are positive, while areas below are negative.

When solving for a definite integral:

• Identify the function \( f(x) \) you want to integrate.
• Determine the interval \([a, b]\), which are the limits of integration.
• Find an antiderivative \( F(x) \) of \( f(x) \).
• Use the Fundamental Theorem of Calculus to compute \( F(b) - F(a) \), which gives the area over the interval.

In the specific exercise we discussed, the function \( f(x) = \frac{1}{\sqrt[3]{x}} \) is integrated over the interval from 8 to 27, resulting in a numerical value of 7.5. This value represents the total area between the curve and the x-axis over the given domain.

The term "antiderivative" refers to a function that reverses differentiation, allowing us to recover a function from its derivative. Given a function \( f(x) \), an antiderivative is any function \( F(x) \) such that \( F'(x) = f(x) \).

Antiderivatives play a crucial role in integration, especially when dealing with definite integrals. When integrating a power function like \( x^n \), the antiderivative is given by the formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), as long as \( n eq -1 \).

In our example, we needed to find the antiderivative of \( x^{-1/3} \). By applying the power rule for integration, the antiderivative can be calculated as:

• Convert the radical expression to a power, so \( f(x) = x^{-1/3} \).
• Integrate using the formula: \( \int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} = \frac{3}{2} x^{2/3} + C \).

This antiderivative is essential for evaluating the definite integral over the specific range from 8 to 27 in the problem.

The Fundamental Theorem of Calculus beautifully bridges the concepts of differentiation and integration. It states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \([a, b]\), then the definite integral of \( f(x) \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).

This theorem highlights two main points:

• It confirms that integration and differentiation are inverse processes.
• It provides a method to evaluate definite integrals easily once the antiderivative is known.

In our exercise, after finding the antiderivative \( F(x) = \frac{3}{2}x^{2/3} \), we applied the Fundamental Theorem of Calculus. We evaluated the antiderivative at the boundaries, substituting 27 for \( x \) and then 8, and calculated the difference: \( F(27) - F(8) \).

This process yielded the area under the curve between \( x = 8 \) and \( x = 27 \), which was computed manually as 7.5. This demonstrates the powerful and practical nature of the Fundamental Theorem of Calculus in solving real-world problems.