Problem 57 - Calculus Single Variable

Question

Calculate the second moment of the given function \(f\) about the vertical axis \(x=c\) for the given \(c\). $$ f(x)=\sin (x) \quad 0 \leq x \leq \pi \quad c=2 \pi $$

Step-by-Step Solution

Answer

The second moment is zero due to symmetry of sine over the interval.

Step 1: Understand the Second Moment Concept

The second moment of a function about a line is defined as the integral of the squared distance of the function from that line. Here, we're asked to find the second moment of the function \( f(x) = \sin(x) \) about the line \( x = c \). This involves computing \( \int_{a}^{b} (x-c)^2 f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

Step 2: Set Up the Integral

We need to set up our integral for the second moment using the given function and interval. The integral becomes:\[\int_{0}^{\pi} (x - 2\pi)^2 \sin(x) \, dx\] where \( c = 2\pi \), \( a = 0 \), and \( b = \pi \).

Step 3: Use Integration by Parts

Since the integral involves \((x - 2\pi)^2 \) and \( \sin(x) \), we can employ the method of integration by parts to simplify it. First, let \( u = (x - 2\pi)^2 \) and \( dv = \sin(x) \, dx \). Then, calculate \( du \) and \( v \).

Step 4: Compute Derivatives and Antiderivatives

Calculate derivatives \( du = 2(x - 2\pi) \, dx \) and the antiderivative \( v = -\cos(x) \). Then, use integration by parts formula \( \int u \, dv = uv - \int v \, du \). Substitute into the formula:\[-(x - 2\pi)^2 \cos(x) \Big|_0^\pi + \int_0^\pi 2(x-2\pi)\cos(x) \, dx\]

Step 5: Integrate Remaining Terms

The integral now needs to be broken down further or approached iteratively with integration by parts again. Work through this process, keeping careful track of signs and values. The resulting computation will give the total second moment.

Step 6: Calculate Definite Integrals and Simplify

Calculate the values of the definite integrals from \(0\) to \(\pi\) using the results from the integration process. Simplification will give the final result for the second moment.

Key Concepts

Integration by Parts

Integration by parts is an important technique in calculus used to integrate products of functions. It is derived from the product rule for differentiation and helps simplify complex integrals by breaking them down into simpler parts.

• To apply integration by parts, first identify two functions: one to differentiate (u'), and one to integrate (dv').
• Use the formula: \[\int u \, dv = uv - \int v \, du\]This expresses the original integral \(\int u \, dv\) in terms of a new integral \(\int v \, du\), which should be easier to solve.
• Careful selection of \(u\) and \(dv\) can determine success. Typically, choose \(u\) as the polynomial part, making the derivative simpler, and \(dv\) as the trigonometric or exponential part.

When working with \((x - 2\pi)^2 \sin(x)\) in our problem, we set \(u = (x - 2\pi)^2\) and \(dv = \sin(x) \, dx\). Carrying out these derivatives and antiderivatives helps progress towards solving the integral successfully.

Definite Integrals

Definite integrals capture the idea of summing up infinitesimal quantities over a specific interval, providing the net area between the function graph and the x-axis.

• Definite integrals have limits of integration, represented as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the bounds or endpoints of integration.
• Evaluating a definite integral involves finding the antiderivative of the function, then computing the difference at the upper and lower boundaries: \[F(b) - F(a)\]
• In our exercise, we compute \(\int_{0}^{\pi} (x - 2\pi)^2 \sin(x) \, dx\), where \(0\) and \(\pi\) are the limits.

The definite integral is essential in many applications, from calculating areas to finding second moments, as demonstrated in this problem.

Sine Function

The sine function, represented as \(\sin(x)\), is a fundamental trigonometric function reflecting waves' periodic nature.

• Its range is between \(-1\) and \(1\), providing a wavelike graph with peaks and troughs.
• The function is periodic with a period of \(2\pi\), meaning it repeats every \(2\pi\) units.
• Key properties of \(\sin(x)\) include its symmetry: it is an odd function, meaning \(\sin(-x) = -\sin(x)\).

In the context of our problem, \(\sin(x)\) acts as part of the integrand, coupling with \((x - 2\pi)^2\) and requiring us to carefully navigate its oscillating properties when integrating, especially over the specified interval from \(0\) to \(\pi\).

Integral Calculus

Integral calculus is the branch of calculus focused on accumulation and area under curves, deeply related to differentiation as its inverse process.

• It is used to find quantities such as area, volume, displacement, and, in this case, second moments.
• Key techniques include substitution, integration by parts, and recognizing integral properties like linearity.
• The fundamental theorem of calculus links differentiation and integration, indicating that differentiation and integration are inverse processes.

In solving the given problem, using integral calculus allows us to calculate the second moment, demonstrating the connection between finding areas and applying these principles to understand functions' behavior over an interval. This practice builds foundational skills crucial for advanced topics in mathematics and engineering.