91Ó°ÊÓ

When calculating the volume of a solid of revolution, using the definite integral is crucial. A definite integral helps us find the accumulation of quantities, such as volume in this case. In the exercise, the definite integral \( \int_{0}^{2} x^{4} \, dx \) is used to accumulate the volumes of smaller slices (or cross-sections) to form the total volume of the solid.

Unlike an indefinite integral, which generalizes the antiderivative of a function, a definite integral evaluates this antiderivative at specific bounds to find a precise value. To perform this task, follow these steps:

• Determine the function to be integrated, which here is the area of a square cross-section, \( x^{4} \).
• Identify the bounds of integration, from 0 to 2, representing the range along the \( x \)-axis.
• Evaluate the definite integral by calculating the antiderivative, and then substituting the upper and lower bounds, using the fundamental theorem of calculus.
• Subtract the result of the lower bound from the upper bound to find the accumulated quantity, resulting in the volume, \( \frac{32}{5} \).

Applying a definite integral like this provides us with the exact amount represented by the solid, showcasing its practical use in finding volumes.

The cross-sectional area is a fundamental concept when dealing with volumes of solids through integration. For solids like the one described in the exercise, the cross-sectional area is constantly changing as it sweeps along the \( x \)-axis.

In the exercise, each cross-section is a square, which makes the direct calculation of its area straightforward. Here's how we approach it:

• Each square's side length is determined by the distance from the curve \( y = x^{2} \) to the \( x \)-axis; effectively, this distance is simply \( x^{2} \).
• Thus, the area \( A \) of each square cross-section, represented by \( A = x^{4} \), is calculated by squaring the side length.
• The function \( A = x^{4} \) then describes how the area changes as \( x \) varies from 0 to 2.

By understanding these changes, one can set up an integral over the specified range to find the total volume. Capturing how these cross-sectional areas sum up along a given interval is crucial for this volume calculation.

Integration techniques are vital for solving problems involving areas and volumes, as illustrated in the exercise. One common technique is determining the antiderivative of a function before applying limits to find the definite integral.

Here are key steps to better understand these techniques:

• Identify the function you need to integrate. In our example, this is \( x^{4} \), representing the area of the square cross-section.
• Determine the antiderivative. For \( x^{4} \), this is \( \frac{x^{5}}{5} \).
• Use the fundamental theorem of calculus: once you have the antiderivative, apply the upper and lower limits of the range (0 to 2 in this case) to find the volume.
• Calculate the difference between these values to get the final result, which in the exercise is \( \frac{32}{5} \).

Mastering these techniques ensures you can handle a variety of integrals, from simple polynomials like \( x^{4} \) to more complex forms. Ensuring you understand each step will enhance your ability to solve similar problems efficiently.