Problem 114 - Calculus Volume 1

Question

In the following exercises, find the average value \(f_{\text { ave }}\) of \(f\) between \(a\) and \(b,\) and find a point \(c,\) where \(f(c)=f_{\text { ave. }}\) \(f(x)=\sin x, a=0, b=2 \pi\)

Step-by-Step Solution

Answer

The average value of \( \sin x \) over \([0, 2\pi]\) is 0, and \(c\) can be 0, \(\pi\), or \(2\pi\).

Step 1: Understand Formula for Average Value

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula: \( f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). We need to apply this formula to find the average value of \(f(x) = \sin x\) between \(a = 0\) and \(b = 2\pi\).

Step 2: Calculate the Integral of the Function

Calculate the definite integral of \( f(x) = \sin x \) from \( 0 \) to \( 2\pi \): \[ \int_{0}^{2\pi} \sin x \, dx = [-\cos x]_{0}^{2\pi} = (-\cos(2\pi)) - (-\cos(0)) = -1 + 1 = 0.\]

Step 3: Find the Average Value

Substitute the result from the integral into the average value formula: \[ f_{\text{ave}} = \frac{1}{2\pi - 0} \times 0 = 0.\]Thus, the average value of \(f(x)\) over \([0, 2\pi]\) is \(0\).

Step 4: Find Point \(c\) Where \(f(c) = f_{\text{ave}}\)

We need to find \(c\) such that \(\sin(c) = 0\). The solutions for this equation on the interval \([0, 2\pi]\) are \(c = 0, \pi, 2\pi\). Thus, any of these values satisfy \(f(c) = f_{\text{ave}}\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus that deals with finding the area under a curve and the accumulation of quantities. It involves two important concepts: the indefinite integral and the definite integral.

• An indefinite integral, often called an antiderivative, represents a family of functions whose derivative gives the original function. It includes a constant of integration, usually denoted as C.
• A definite integral calculates the total accumulation of values between two specified endpoints on a function, often representing the area under the curve.

In the given exercise, we use the concept of a definite integral. The integral \(\int_{0}^{2\pi} \sin x \, dx \) helps in determining the accumulation of the function \(\sin x\) over the interval from 0 to \(2\pi\). Understanding to integrate trigonometric functions like sine enables us to determine average values and solve real-world problems involving periodic phenomena.

Trigonometric Functions

Trigonometric functions are essential in mathematics, especially when dealing with periodic functions such as waves and oscillations. These functions include sine, cosine, and tangent, among others.

• The sine function, denoted by \(\sin x\), describes the y-coordinate of a point on the unit circle as the angle x varies.
• Trigonometric functions like sine are periodic, with sine having a period of \(2\pi\), which means the function repeats every \(2\pi\) units.

In the problem, we focus on the sine function, \(f(x) = \sin x\). Its periodicity is key, as it helps find values like average value over full cycles, seamlessly repeating its behavior. Solving \(\sin(c) = 0\) lets us find specific angles where the sine equals zero, confirming the periodic behavior.

Definite Integral

A definite integral is a powerful tool in calculus used to find areas, volumes, central points, and many other analytical properties. The integral has limits and gives a number, representing the cumulative value of a function from one point to another.

• Set up with lower and upper limits, such as \(a\) and \(b\), it is expressed as \([ \int_{a}^{b} f(x) \, dx ]\).
• The definite integral finds net areas; regions above x-axis are positive, while those below are negative.

For the function \( \sin x \,\), calculating its definite integral between \(0\) and \(2\pi\) results in zero, due to symmetrical areas canceling each other out. This understanding is crucial for evaluating the average, where the whole interval's contribution sums to zero, simplifying the process of finding average values over aligned intervals.