Chapter 5: Problem 10
Find the average value of the function over the given interval. $$f(x)=\frac{1}{\sqrt{1-x^{2}}} ;\left[-\frac{1}{2}, 0\right]$$
Short Answer
Expert verified
The average value is \( \frac{\pi}{3} \).
Step by step solution
01
Identify the Formula for Average Value
The formula for the average value of a continuous function \( f(x) \) over the interval \([a, b]\) is given by: \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). Our function is \( f(x) = \frac{1}{\sqrt{1-x^{2}}} \), and the interval is \( \left[-\frac{1}{2}, 0\right] \).
02
Set up the Integral
Substitute the given function and interval into the average value formula: \( \frac{1}{0 - \left(-\frac{1}{2}\right)} \int_{-\frac{1}{2}}^{0} \frac{1}{\sqrt{1-x^{2}}} \, dx \). This simplifies to \( 2 \int_{-\frac{1}{2}}^{0} \frac{1}{\sqrt{1-x^{2}}} \, dx \).
03
Compute the Integral
The integral \( \int \frac{1}{\sqrt{1-x^{2}}} \, dx \) is a standard form that is equal to \( \sin^{-1}(x) + C \). Compute the definite integral from \(-\frac{1}{2}\) to \(0\):\[ \int_{-\frac{1}{2}}^{0} \frac{1}{\sqrt{1-x^{2}}} \, dx = \sin^{-1}(0) - \sin^{-1}\left(-\frac{1}{2}\right) \].
04
Evaluate the Trigonometric Functions
Evaluate \( \sin^{-1}(0) \) and \( \sin^{-1}\left(-\frac{1}{2}\right) \): \( \sin^{-1}(0) = 0 \) and \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \). Thus the evaluation of the integral becomes:\[ 0 - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} \].
05
Calculate the Average Value
Insert the result of the integral back into the average value formula: \( 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \). This is the average value of the function over the interval \( \left[-\frac{1}{2}, 0\right] \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a core component in calculus that help us calculate the accumulation of quantities, such as area under a curve, over a specified interval. In simple terms, a definite integral gives us the total size (area, in most cases) of what is being measured over a specific section of the function's domain. For example, when we have a function like \( f(x) \) and we want to find its average value over a certain interval \([a, b]\), a definite integral helps us determine the area under the function's graph between these two points.
To find the average value, we use the formula:
To find the average value, we use the formula:
- \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)
Inverse Trigonometric Functions
Inverse trigonometric functions are extremely useful when dealing with integrals like our current example. Expressions such as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \) are inverse functions of the standard sine, cosine, and tangent functions. These are also sometimes called arcsine, arccosine, and arctangent, respectively. Here's a quick overview:
- \( \sin^{-1}(x) \) refers to the angle whose sine is \( x \)
- \( \cos^{-1}(x) \) refers to the angle whose cosine is \( x \)
- \( \tan^{-1}(x) \) refers to the angle whose tangent is \( x \)
Interval Notation
Interval notation is a way of writing subsets of the real number line. It's a concise way of expressing a range of values for which a function is to be considered. In mathematics, it is essential to clearly understand and denote the intervals over which functions are to be analyzed or graphed. Here's a quick guide:
- \([a, b]\): Closed interval where \(a\) and \(b\) are included in the set
- \((a, b)\): Open interval where \(a\) and \(b\) are not included
- \([a, b)\): \(a\) is included, \(b\) is not
- \((a, b]\): \(a\) is not included, \(b\) is