91Ó°ÊÓ

Continuous functions are fundamental in mathematics because they have no breaks, jumps, or holes in their graphs. This means that for a function \( f \) defined on an interval \([a, b] \), the function does not suddenly jump from one value to another; every value in between is covered smoothly.
In our exercise, the function \( f \) is continuous on the interval \([0, \, \infty)\). This property ensures that all intermediate values are taken without any sudden jumps, which is essential when performing operations like integration and volume calculations.
For instance, since \( f \) is continuous and increasing, \( f^2(x) \) is also continuous. This is critical for our problem because it allows us to apply techniques like the Intermediate Value Theorem efficiently.

Integral calculus allows us to calculate areas under curves, which is crucial for finding the volume of solids. In this problem, we use integral calculus to find the volume generated by revolving a function \( f(x)\) around the x-axis.
We start by setting up the integral for the volume under the function. The formula is:
\[ V = \int_0^b \pi [f(x)]^2 \, dx \]
Here, \( \pi [f(x)]^2 \) represents the area of a circular cross-section of the solid at a point \( x \), and integrating this from 0 to \( b \) sums up all these infinitesimally small volumes to give the total volume.
Similarly, for the volume of the cylinder formed by rotating \(y = f(b)\) around the x-axis, we use:
\[ V_c = \pi [f(b)]^2 b \]
Understanding both of these integrals is key to solving the problem. We equate the volume under the function to half the volume of the cylinder and solve for \( b \).

The Intermediate Value Theorem (IVT) is an essential concept in calculus. It states that if a function \( f \) is continuous on a closed interval \([a, b]\), then for any value \( L \) between \( f(a) \) and \( f(b) \), there exists a number \( c \in [a, b] \) such that \( f(c) = L \).
In this exercise, we leverage the IVT to show that there exists a value \( b > 0 \) for which our equation
\[ \int_0^b [f(x)]^2 \, dx = \frac{1}{2} [f(b)]^2 b \]
holds. Since \( f(x) \) is continuous and increasing, \( f^2(x) \) is continuous, making the integral on the left a continuous function of \( b \).
The right-hand side of the equation is also continuous and non-decreasing because \( [f(b)]^2 b \) increases as \( b \) increases. Therefore, by IVT, there must be some \( b > 0 \) that satisfies the equation.

Finding the volume of solids of revolution involves rotating a curve around an axis and calculating the resulting 3D shape's volume. For a function \( f(x)\) rotated around the x-axis, the volume element at point \( x \) is a disk with radius \( f(x) \) and thickness \( dx \).
The volume of this disk is \( \pi [f(x)]^2 dx \). Summing these disk volumes from \( x = 0 \) to \( x = b \) using an integral gives:
\[ V = \int_0^b \pi [f(x)]^2 \, dx \]
Similarly, a cylinder formed by rotating the line segment \( y = f(b) \), where \( 0 \leq x \leq b \), around the x-axis has a constant radius \( f(b) \) and height \( b \). Its volume is calculated as:
\[ V_c = \pi [f(b)]^2 b \]
In this exercise, we show that there is a point \( b \) where the volume under \( f(x) \) rotated around the x-axis is half that of the cylinder’s volume.