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Magazine Chapter 28: Problem 35
Solve the given problems by integration. Find the root-mean-square value of the function \(y=\sqrt{\sin ^{-1} x}\) between \(x=0 \text { and } x=1 . \text { (See Example } 7 \text { of Section } 28.5 .).\)
Short Answer
Expert verified
The RMS value of \(y = \sqrt{\sin^{-1} x}\) from \(x=0\) to \(x=1\) is \(\sqrt{\frac{\pi}{2}}\).
Step by step solution
01
Understand the problem
We need to find the root-mean-square (RMS) value of the function \(y = \sqrt{\sin^{-1} x}\) over the interval from \(x = 0\) to \(x = 1\). RMS value for a function \(f(x)\) from \(a\) to \(b\) is \(\sqrt{\frac{1}{b-a} \int_a^b [f(x)]^2 \, dx}\).
02
Set up the integral
Square the function to get the integral of \((\sin^{-1} x)\). The formula for RMS becomes \(\sqrt{\frac{1}{1-0} \int_0^1 (\sin^{-1} x) \, dx}\).
03
Integrate \((\sin^{-1} x)\)
The integral of \(\sin^{-1} x\) with respect to \(x\) requires integration by parts. Set \(u = \sin^{-1} x\) and \(dv = dx\). The derivative \(du = \frac{1}{\sqrt{1-x^2}} dx\) and \(v = x\). Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
04
Calculate the integral value
Following integration by parts, we get \[x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} \, dx\]. Solve the remaining integral. Substitute to solve \(\int \frac{x}{\sqrt{1-x^2}} \, dx\) by letting \(w = 1-x^2\), giving the remaining integral as \(-\sqrt{1-x^2}\). Evaluate from 0 to 1.
05
Solve definite integral
Merge back the solution: \[x \sin^{-1} x + \sqrt{1-x^2} \bigg|_0^1\]. Calculating at these limits, at \(x=1\), the value is \(\frac{\pi}{2}\). At \(x=0\), it is zero. The definite integral evaluates to \(\frac{\pi}{2}\).
06
Compute RMS value
With the integral evaluated as \(\frac{\pi}{2}\), apply it to the RMS formula: \(\sqrt{\int_0^1 (\sin^{-1} x) \, dx}\) which simplifies to \(\sqrt{\frac{\pi}{2}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Root-Mean-Square Value
The root-mean-square (RMS) value is a way to determine the average of a function over a certain interval. It is similar to the average but takes into account the square of the function values, offering a measure of the function's magnitude. To find the RMS value of a function like \(y = \sqrt{\sin^{-1} x}\), follow these steps:
- Square the function: To compute the average of its squares over an interval.
- Integrate the squared function: Calculate the area under its curve between a certain range.
- Divide by the range: Normalize the area to get the average of squares.
- Take the square root: Convert the result back from a squared form to the RMS value.
Integration by Parts
Integration by parts is a powerful technique for integrating products of functions. It is derived from the product rule for differentiation and is given by the formula: \(\int u \, dv = uv - \int v \, du\). Here's how you can apply it:
- Select \(u\) and \(dv\): Choose parts of your integrand to differentiate (\(du\)) and integrate (\(v\)). The choice can affect simplification, so picking \(u\) as the part that simplifies upon differentiation is usually helpful.
- Differentiation: Find \(du\) by differentiating \(u\).
- Integration: Find \(v\) by integrating \(dv\).
- Apply the formula: Substitute into the integration by parts equation and solve.
Definite Integral
A definite integral calculates the net area between a function and the x-axis over a specific interval \([a, b]\). Mathematically, it is represented as \(\int_a^b f(x) \, dx\). Here鈥檚 a quick guide on definite integrals:
- Set Boundaries: Define the limits \(a\) and \(b\), which indicate where on the x-axis the function should be evaluated.
- Compute Indefinite Integral: Find the antiderivative of the function, represented as \(F(x)\).
- Evaluate the Limits: Substitute the upper and lower bounds into \(F(x)\) and subtract \(F(a)\) from \(F(b)\) to get the net area.
Inverse Trigonometric Function
Inverse trigonometric functions, like \(\sin^{-1} x\), serve as the inverse of the standard trigonometric functions. They return the angle whose sine, cosine, or tangent equals a given number. Here's what makes them special in calculus:
- Range Limitations: Inverse trig functions are limited to specific ranges to maintain a one-to-one correspondence, ensuring each output derives from a single input.
- Differentiability: Each has a unique derivative, e.g., \(\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\), crucial in solving integration problems by substitution or parts.
- Applications: Used in calculus to solve integrals involving arc lengths or angles.
