Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all values of \(c\) that satisfy the Mean Value Theorem for Integrals on the given interval. $$ f(x)=x^{2} ; \quad[-1,1] $$

Step-by-Step Solution

Answer

The values of \(c\) are \(\pm \frac{1}{\sqrt{3}}\).

Step 1: Understand the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\), then there exists at least one number \(c\) in the interval \([a, b]\) such that \(f(c)(b-a) = \int_{a}^{b} f(x) \, dx\). Here, \(f(x) = x^2\) on \([-1, 1]\).

Step 2: Calculate the Definite Integral

We need to compute the definite integral \(\int_{-1}^{1} x^2 \, dx\). Use the power rule for integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Here, \(\int x^2 \, dx = \frac{x^3}{3} + C\). Evaluating at the bounds, you get \(\frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}\).

Step 3: Apply the Formula from the Theorem

According to the theorem, \(f(c)(b-a) = \int_{-1}^{1} x^2 \, dx\). Substitute the known values: \(f(c) \times (1 - (-1)) = \frac{2}{3}\). So \(f(c) \times 2 = \frac{2}{3}\).

Step 4: Solve for \(c\)

We need \(f(c) = c^2\) so that \(c^2 \times 2 = \frac{2}{3}\). Solve for \(c^2\) by dividing both sides by 2: \(c^2 = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}\). Take the square root of both sides: \(c = \pm \frac{1}{\sqrt{3}}\).

Step 5: Verify \(c\) is in the Interval

Check if \(c = \frac{1}{\sqrt{3}}\) and \(c = -\frac{1}{\sqrt{3}}\) lie within the interval \([-1, 1]\). Since both values are between \(-1\) and \(1\), they are valid.

Key Concepts

Continuous Function

A continuous function is one where small changes in the input result in small changes in the output. In other words, there are no jumps, breaks, or holes in the graph of the function. It can be visualized as a smooth, unbroken curve.In the context of the Mean Value Theorem for Integrals, having a continuous function is essential. This ensures that the definite integral value and the corresponding mean value point, where the average rate occurs, exist without discrepancies. When we say a function like \(f(x) = x^2\) is continuous on the interval

• \([-1, 1]\), it means that for every value within this interval, the function is continuous.
• No sudden jumps or holes interrupt the flow of the curve between \(-1\) and \(1\).

Definite Integral

A definite integral is the evaluation of the integral of a function over a specified interval. It represents the accumulated area under the curve of a function from the lower limit to the upper limit of integration.In the problem, the definite integral was calculated using the power rule of integration, which is a method for finding antiderivatives. The integral \(\int_{-1}^{1} x^2 \, dx\) measures the total area under the curve \(f(x)=x^2\) from \(x = -1\) to \(x = 1\). The value we obtained is \(\frac{2}{3}\) after solving the integral using the steps:

• Apply the power rule formula: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\).
• Calculate the antiderivative: \(\int x^2 \, dx = \frac{x^3}{3} + C\).
• Evaluate at the bounds: \(\frac{1^3}{3} - \left(-\frac{1^3}{3}\right) = \frac{2}{3}\).

This tells us the approximate total area under the curve \(x^2\) from \(-1\) to \(1\).

Power Rule for Integration

The power rule for integration is a fundamental rule used to find the integral of functions that are power functions. It's a straightforward method that allows us to move from finding an antiderivative to solving integrals.The formula for the power rule is \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(n\) is not equal to \(-1\). Here's how it works:- Increment the exponent \(n\) by one.- Divide by the new exponent.- Don't forget to introduce the constant of integration \(C\) when finding the general antiderivative.In our example, the power rule enabled us to find the antiderivative of \(x^2\), using:

• The power of \(x\) increased from \(2\) to \(3\).
• The expression became \(\frac{x^3}{3} + C\).

This specific rule helps simplify the process of integration, especially in polynomial functions.

Closed Interval

A closed interval in mathematics is an interval where the endpoints are included. It's usually denoted by square brackets. For instance, \([-1, 1]\) is a closed interval where both \(-1\) and \(1\) are part of the interval.Closed intervals are significant in calculus, particularly when using the Mean Value Theorem for Integrals.

• The theorem requires that the function is continuous over a closed interval.
• The value \(c\), in the theorem, must lie within this closed interval.

This ensures that the outcomes of calculations and properties used are valid for the endpoints as well as any points in between. When verifying if values like \(c = \frac{1}{\sqrt{3}}\) and \(c = -\frac{1}{\sqrt{3}}\) satisfy the Mean Value Theorem, confirming they are within the closed interval \([-1, 1]\) is crucial. This assures they are valid solutions in our example.