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Chapter 4: Problem 39

Find the average value over the given interval. $$y=4-x^{2} ; \quad[-2,2]$$

Short Answer

Expert verified
The average value is \( \frac{8}{3} \).

Step by step solution

01

- Understand the Function and Interval

Identify the function to be evaluated and the interval over which to find the average value. Here, the function is given by \( y = 4 - x^2 \) and the interval is \([-2, 2]\).
02

- Average Value Formula

Recall the formula for the average value of a function \(f(x)\) over the interval \([a, b]\): \[ \text{Average value} = \frac{1}{b-a} \times \text{Area under } f(x) \text{ from } a \text{ to } b. \] In this case, \(a = -2\) and \(b = 2\).
03

- Calculate the Length of the Interval

Determine the length of the interval \( [a, b] \) which is given by \( b - a \): \[ 2 - (-2) = 4. \]
04

- Set Up the Integral

Set up the integral to find the area under the curve of the function. The integral of \( y = 4 - x^2 \) from \( -2 \) to \( 2 \) is: \[ \text{Area} = \frac{1}{4} \times \text{Integral} \bigg[ \text{from } -2 \text{ to } 2 \bigg] (4 - x^2) dx. \]
05

- Compute the Integral

Compute the integral value: \[ \text{Integral} \bigg[ \text{from } -2 \text{ to } 2 \bigg] (4 - x^2) dx = \bigg[ 4x - \frac{x^3}{3} \bigg] \bigg|_{-2}^{2} = \bigg[ 4(2) - \frac{(2)^3}{3} \bigg] - \bigg[ 4(-2) - \frac{(-2)^3}{3} \bigg]. \]
06

- Simplify the Integral Result

Simplify the integral result: \[ = \bigg[ 8 - \frac{8}{3} \bigg] - \bigg[ -8 + \frac{8}{3} \bigg] = \bigg[ \frac{24}{3} - \frac{8}{3} \bigg] - \bigg[ -\frac{24}{3} + \frac{8}{3} \bigg] = \bigg[ \frac{16}{3} \bigg] + \bigg[ \frac{16}{3} \bigg] = \frac{32}{3}. \]
07

- Calculate the Average Value

Divide the integral result by the length of the interval \(4\): \[ \text{Average value} = \frac{1}{4} \times \frac{32}{3} = \frac{32}{12} = \frac{8}{3}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a major branch of calculus that focuses on the accumulation of quantities and the areas under and between curves. There are two main types of integrals: definite and indefinite.
In this problem, we are dealing with the concept of finding the average value of a function over a specified interval. This requires us to calculate a definite integral.
Definite Integral
A definite integral represents the area under a curve within a certain interval. It's denoted as \(\text{Integral} \big[ a \text{ to } b \big] f(x) \text{ d}x\), where \([a, b]\) is the interval, and \ f(x) \ is the function.
In our solution, we've set up the integral from \(-2 \text{ to } 2\)\to find the area under the curve for the function \ y = 4 - x^2\. By solving this integral, we calculated the total area under the curve.
Average Value Theorem
The Average Value Theorem for integrals helps us find the average value of a function over a specific interval. The average value of a function \( f(x) \) on an interval \([a, b]\) is given by:
\( \text{Average value} = \frac{1}{b-a} \times \text{Integral} [a \text{ to } b] f(x) \text{ d}x \).
In our example, we applied this theorem to the function \ y = 4 - x^2 \ over the interval \ [-2, 2] \.By calculating the definite integral and dividing it by the length of the interval, we found the average value to be \ \frac{8}{3} \.

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