Problem 52 - Calculus Single Variable

Question

In each of Exercises \(43-52\) calculate the average of the given expression over the given interval. $$ 24 x \arcsin (x) \quad 0 \leq x \leq 1 / 2 $$

Step-by-Step Solution

Answer

The average value of the function over \([0, \frac{1}{2}]\) is \( \frac{\pi}{8} \).

Step 1: Understand the Problem

We have to find the average value of the function \( f(x) = 24x \arcsin(x) \) over the interval \([0, \frac{1}{2}]\). The average value of a continuous function \( f \) on an interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_a^b f(x) \, dx \).

Step 2: Set Up the Integral

First, identify the interval from \( a = 0 \) to \( b = \frac{1}{2} \). The formula for the average value becomes \( \frac{1}{\frac{1}{2} - 0} \int_0^{\frac{1}{2}} 24x \arcsin(x) \, dx \). This simplifies to \( 2 \int_0^{\frac{1}{2}} 24x \arcsin(x) \, dx \).

Step 3: Simplify the Integral Expression

Factor out the constant term from the integral: \( 48 \int_0^{\frac{1}{2}} x \arcsin(x) \, dx \). We will solve this integral using integration by parts, where \( u = \arcsin(x) \) and \( dv = x \, dx \).

Step 4: Integration by Parts Setup

Let \( u = \arcsin(x) \). Then \( du = \frac{1}{\sqrt{1-x^2}} \, dx \). Let \( dv = x \, dx \), so \( v = \frac{x^2}{2} \). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 5: Apply Integration by Parts

Substitute into the formula: \( uv = \arcsin(x) \cdot \frac{x^2}{2} \) from \(0\) to \(\frac{1}{2}\), and \( \int v \, du = \int \frac{x^2}{2} \cdot \frac{1}{\sqrt{1-x^2}} \, dx \). Evaluate the expression for \(uv\) first.

Step 6: Evaluate uv from 0 to 1/2

Calculate \( \left[ \arcsin(x) \cdot \frac{x^2}{2} \right]_0^{1/2} \). This gives us \( \left( \arcsin(\frac{1}{2}) \cdot \frac{(\frac{1}{2})^2}{2} \right) - \left( \arcsin(0) \cdot 0 \right) = \left( \frac{\pi}{6} \cdot \frac{1}{8} \right) \).

Step 7: Simplify and Solve the Remaining Integral

Now solve \( \int \frac{x^2}{2\sqrt{1-x^2}} \, dx \). This integral is more complex but can be solved using trigonometric substitution or known integral tables. After obtaining the result, multiply by 48 as per Step 3.

Step 8: Combine Results and Compute Final Average

Combine the results from \(uv\) and the solved integral part to find the complete \( \int_0^{1/2} x \arcsin(x) \, dx \), followed by multiplying by 48. The final output gives the average value.

Key Concepts

Integration by Parts

Integration by parts is a powerful method used to solve integrals where the standard methods might not be sufficient. It is particularly useful when dealing with products of functions. The key formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here's how you can think about applying this method:

• **Identify the components**: Choose parts of your function to be \( u \) and \( dv \). Typically, \( u \) is chosen to be a function that simplifies when differentiated, while \( dv \) is chosen to be the remaining part of the product that can be easily integrated.
• **Differentiate and Integrate**: Determine \( du \) by differentiating \( u \), and \( v \) by integrating \( dv \).
• **Substitute in the formula**: Plug these values into the integration by parts formula. This transforms the original integral into a potentially simpler problem to solve.

In the context of our problem, integration by parts applies with \( u = \arcsin(x) \) and \( dv = x \, dx \). Differentiating \( u \) and integrating \( dv \) respectively yields \( du = \frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = \frac{x^2}{2} \). By substituting these into the integration by parts formula, we simplify the problem to a product minus another integral.

Definite Integral

The definite integral is a concept that allows us to calculate the accumulation of quantities, such as area under a curve, over an interval \([a, b]\). Its standard form is:

• \( \int_a^b f(x) \, dx \)

The process of computing a definite integral involves:

• **Finding the antiderivative**: First, determine the indefinite integral or antiderivative, \( F(x) \), of the function \( f(x) \).
• **Evaluating the limits**: Substitute the upper and lower bounds into the antiderivative: \( F(b) - F(a) \).

In our example, the definite integral \( \int_0^{1/2} 24x \arcsin(x) \, dx \) is set up by first identifying the limits from \( a = 0 \) to \( b = \frac{1}{2} \). After solving with integration by parts, the definite integral process allows us to determine the exact area under the curve from 0 to \( \frac{1}{2} \), which forms the foundation for calculating the average value of the function over that interval.

Arcsin Function

The arcsin function, also written as \( \arcsin(x) \), is the inverse of the sine function, specifically for angles in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). It answers the question: "What angle \( \theta \) has a sine of \( x \)?"

• **Properties**: The arcsin function takes an input value, \( x \), restricted between -1 and 1, and produces an output angle in radians.
• **Differentiation**: The derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \), which is often crucial for calculus problems involving trigonometric inverse functions.

In the original exercise, \( \arcsin(x) \) plays a core role in forming the product \( 24x \arcsin(x) \). When using integration by parts, understanding the properties of the arcsin function and its derivative helps in choosing \( u = \arcsin(x) \) efficiently, which simplifies the integral considerably compared to other potential choices.