Problem 24 - Calculus

Question

Find all values of c that satisfy the Mean Value Theorem for Integrals on the given interval. $$ T(x)=x^{3} ; \quad[0,2] $$

Step-by-Step Solution

Answer

The value of \( c \) is \( \sqrt[3]{2} \approx 1.26 \).

Step 1: Understand the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( T(x) \) is continuous on the closed interval \([a, b]\), there exists at least one number \( c \) in the open interval \((a, b)\) such that: \[T(c) \times (b-a) = \int_{a}^{b} T(x) \, dx\] We need to find such a \( c \) for the function \( T(x) = x^3 \) on the interval \([0, 2]\).

Step 2: Calculate the Definite Integral

Calculate the integral of \( T(x) = x^3 \) over the interval \([0, 2]\):\[\int_{0}^{2} x^3 \, dx\]The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \). Evaluate this from 0 to 2:\[\left. \frac{x^4}{4} \right|_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4\]

Step 3: Set Up the Mean Value Equation

Using the result of the integral, set up the equation from the theorem using \( c \), where \( T(c) = c^3 \) and \( b-a = 2-0 = 2 \):\[c^3 \times 2 = 4\]Simplify to find \( c^3 \):\[c^3 = 2\]

Step 4: Solve for c

Now, solve \( c^3 = 2 \) to find \( c \):\[c = \sqrt[3]{2}\]Thus, \( c = \sqrt[3]{2} \approx 1.26 \) (approximately). This \( c \) is within the interval \((0, 2)\).

Key Concepts

Continuous Function

A continuous function is a type of function where, intuitively, you can draw its graph without lifting your pen. More precisely, it's a function that does not have any jumps, breaks, or holes in its graph.
For a formal definition, a function \( f(x) \) is considered continuous at a point \( a \) if the following three conditions are met:

• The function \( f(a) \) is defined.
• The limit of \( f(x) \) as \( x \) approaches \( a \) from both sides exists.


• The limit equals \( f(a) \).

Continuous functions are crucial in calculus, specifically in the application of the Mean Value Theorems, because these theorems require the function to be continuous over a closed interval They ensure that the behavior of the function remains predictable and smooth across that interval.

Definite Integral

The definite integral of a function \( f(x) \) over a closed interval \([a, b]\) is a fundamental concept in calculus. It represents the net "area" under the curve of the function from \( x=a \) to \( x=b \).
Mathematically, it is denoted as: \[ \int_{a}^{b} f(x) \, dx \]

• The actual process involves calculating an antiderivative of the function, also known as the indefinite integral.


• The antiderivative is then evaluated at the two endpoints \( a \) and \( b \).


• Finally, you subtract the result of substituting \( a \) from that of \( b \) in the antiderivative.

For example, in the function \( T(x) = x^3 \), the definite integral over \([0, 2]\) allows us to find the total area under the cubic curve over that interval, which is crucial for applying the Mean Value Theorem for Integrals.

Closed Interval

In calculus, a closed interval is a set of real numbers that includes all numbers between two endpoints, including the endpoints themselves. It is denoted as \([a, b]\).

• "Closed" means that the interval contains its boundary points \( a \) and \( b \).


• Graphically, the endpoints are included and represented as filled dots.
• The inclusion of the endpoints is significant when checking continuity for the application of certain theorems, such as the Mean Value Theorem for Integrals.

For the function \( T(x) = x^3 \), the closed interval \([0, 2]\) ensures the function is being evaluated over a specific range that includes 0 and 2, allowing application of integration techniques and fulfilling the conditions required to apply the Mean Value Theorem for Integrals.

Cubic Function

A cubic function is a polynomial function of the form \( f(x) = ax^3 + bx^2 + cx + d \), where \( a eq 0 \). This function is characterized by its degree of 3, meaning the highest power of \( x \) is 3.

• Cubic functions have at most three real roots (or solutions) and can have up to two turning points.


• They typically have an "S" shaped curve with one inflection point, where the concavity changes.
• The cubic function \( T(x) = x^3 \) used in the exercise is a simple form without subsidiary x, squared, or constant term coefficients.

This type of function smoothly and continuously increases or decreases over its domain, making it an excellent candidate for applying the Mean Value Theorem for Integrals.
In the given problem, understanding the nature of the cubic function is essential for evaluating the definite integral and identifying points that satisfy the theorem.