Problem 33
Question
Find a value of \(c\) that satisfies the conclusion of the Integral Mean Value Theorem. $$\int_{0}^{2} 3 x^{2} d x(=8)$$
Step-by-Step Solution
Verified Answer
The value of \(c\) that satisfies the conclusion of the Integral Mean Value Theorem for the given function and limits is \(c = \sqrt{\frac{4}{3}}\).
1Step 1: Evaluate Integral
First, we must evaluate the integral from \(a\) to \(b\) for the function \(3x^{2}\). Performing the calculations, we find that \(\int_{0}^{2} 3 x^{2} dx = 8\) as provided in the exercise. Thus, the area under the curve \(3x^2\) between \(x=0\) and \(x=2\) is 8.
2Step 2: Apply Integral Mean Value Theorem
Now, applying the Integral Mean Value Theorem, we set up the equation \((b-a)f(c) = \int_{a}^{b} f(x) dx\). Substituting the given values, we get \((2-0)f(c) = 8\). This simplifies to \(2f(c) = 8\).
3Step 3: Solve for \(f(c)\)
Solving for \(f(c)\), we have \(f(c) = 8/2 = 4\). However, our function \(f(x)\) is \(3x^2 \), which means that \(3c^2 = 4\).
4Step 4: Solve for \(c\)
Finally, to find the value of \(c\), we can solve the equation \(3c^2 = 4\) for \(c\). Doing so gives \(c = \sqrt{\frac{4}{3}}\) or \(c = -\sqrt{\frac{4}{3}}\). However, as \(c\) must lie in the interval \([0,2]\), we discard the negative root and finally find \(c = \sqrt{\frac{4}{3}}\).
Key Concepts
Definite IntegralFundamental Theorem of CalculusMean Value Theorem for Integrals
Definite Integral
Understanding the concept of a definite integral is crucial for a firm grasp of calculus. In its essence, the definite integral evaluates the accumulated sum of areas under the curve of a function between two points, usually referred to as the bounds or limits of integration. The process of finding this integral is known as integration, and it can be visualized as the total area under the curve of a function f(x), from the start point a to the end point b.
For instance, let's consider the definite integral of the function 3x^2 between 0 and 2, denoted as \( \int_{0}^{2} 3 x^{2} dx \). In this exercise, it is given that the value of this integral is 8, meaning that the area under the curve 3x^2 from x=0 to x=2 is 8 square units.
To calculate a definite integral, one typically uses the Fundamental Theorem of Calculus or integration techniques like substitution, by parts, or numerical methods when an antiderivative is difficult to establish. The power rule for integration, for example, would be applicable in this case because 3x^2 is a power function.
For instance, let's consider the definite integral of the function 3x^2 between 0 and 2, denoted as \( \int_{0}^{2} 3 x^{2} dx \). In this exercise, it is given that the value of this integral is 8, meaning that the area under the curve 3x^2 from x=0 to x=2 is 8 square units.
To calculate a definite integral, one typically uses the Fundamental Theorem of Calculus or integration techniques like substitution, by parts, or numerical methods when an antiderivative is difficult to establish. The power rule for integration, for example, would be applicable in this case because 3x^2 is a power function.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone linking the concept of differentiation and integration. It comprises two parts. The first part ensures that if a function f(x) is continuous over an interval [a, b], then the function has an antiderivative within that interval. This antiderivative can be used to calculate definite integrals.
The second part states that the definite integral from a to b of a function can be found by evaluating the antiderivative at the upper limit b and subtracting its evaluation at the lower limit a. Represented symbolically, if F(x) is the antiderivative of f(x), then \( \int_{a}^{b} f(x) dx = F(b) - F(a) \).
Applying this theorem makes it possible to evaluate the integral from the previous example, and this relationship is the foundation on which the calculation in the exercise is built. When we found that \( \int_{0}^{2} 3 x^{2} dx = 8 \), we implicitly relied on the Fundamental Theorem of Calculus.
The second part states that the definite integral from a to b of a function can be found by evaluating the antiderivative at the upper limit b and subtracting its evaluation at the lower limit a. Represented symbolically, if F(x) is the antiderivative of f(x), then \( \int_{a}^{b} f(x) dx = F(b) - F(a) \).
Applying this theorem makes it possible to evaluate the integral from the previous example, and this relationship is the foundation on which the calculation in the exercise is built. When we found that \( \int_{0}^{2} 3 x^{2} dx = 8 \), we implicitly relied on the Fundamental Theorem of Calculus.
Mean Value Theorem for Integrals
The Mean Value Theorem for Integrals is a concept that bridges the average value of a function over an interval with a specific value the function takes within that interval. In simpler terms, it guarantees that for a continuous function f(x) over the closed interval [a, b], there exists at least one point c in (a, b) such that f(c) is equal to the average value of f(x) over [a, b].
Mathematically, it is expressed by the equation \( (b-a)f(c) = \int_{a}^{b} f(x) dx \). In this instance, we are given that the interval is from 0 to 2, and the definite integral of 3x^2 over this interval is 8. Using the theorem, we established the equation \( 2f(c) = 8 \), signifying that the average value of the function over this interval is 4. As concluded from the exercise, since f(x) = 3x^2, we find that \(c\) equals \( \sqrt{\frac{4}{3}} \) by solving \( 3c^2 = 4 \). This theorem solidifies our understanding that no matter the function's shape, as long as it's continuous, a 'representative' value or average, so to speak, can be pinpointed within its domain of definition.
Mathematically, it is expressed by the equation \( (b-a)f(c) = \int_{a}^{b} f(x) dx \). In this instance, we are given that the interval is from 0 to 2, and the definite integral of 3x^2 over this interval is 8. Using the theorem, we established the equation \( 2f(c) = 8 \), signifying that the average value of the function over this interval is 4. As concluded from the exercise, since f(x) = 3x^2, we find that \(c\) equals \( \sqrt{\frac{4}{3}} \) by solving \( 3c^2 = 4 \). This theorem solidifies our understanding that no matter the function's shape, as long as it's continuous, a 'representative' value or average, so to speak, can be pinpointed within its domain of definition.
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