Problem 6 - Calculus Single Variable

Question

In each of Exercises \(1-12,\) calculate the average value of the given function on the given interval. $$ f(x)=1+\cos (x) \quad I=[0,2 \pi] $$

Step-by-Step Solution

Answer

The average value of the function is 1.

Step 1: Understand the Problem

The problem asks for the average value of the function \( f(x)=1+\cos(x) \) on the interval \( I=[0,2\pi] \). The average value of a function \( f(x) \) over an interval \([a,b]\) is given by the formula: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \; dx \].

Step 2: Set Up the Integral

To find the average value \( f_{avg} \), we need to compute the integral \( \int_{0}^{2\pi} (1+\cos(x)) \; dx \). The length of the interval \([0, 2\pi]\) is \(2\pi - 0 = 2\pi\). Therefore, the average value is given by: \[ \frac{1}{2\pi} \int_{0}^{2\pi} (1+\cos(x)) \; dx \].

Step 3: Compute the Integral

First, calculate the integral: \( \int_{0}^{2\pi} (1+\cos(x)) \; dx = \int_{0}^{2\pi} 1 \; dx + \int_{0}^{2\pi} \cos(x) \; dx \). The integral of 1 over \([0,2\pi]\) is \(2\pi\). The integral of \(\cos(x)\) is \(\sin(x)\), so \(\int_{0}^{2\pi} \cos(x) \; dx = \sin(2\pi) - \sin(0) = 0 - 0 = 0\). Thus, \(\int_{0}^{2\pi} (1+\cos(x)) \; dx = 2\pi + 0 = 2\pi\).

Step 4: Calculate the Average Value

Now, substitute the result of the integral into the average value formula: \[ f_{avg} = \frac{1}{2\pi} \times 2\pi = 1 \].

Step 5: Conclude the Solution

The average value of the function \( f(x) = 1 + \cos(x) \) over the interval \([0, 2\pi]\) is \( 1 \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental aspect of calculus that concerns itself with the concept of integration. At its core, integration is the process of finding the integral of a function. It helps in calculating areas under curves, volumes, and more. In the context of finding the average value of a function across an interval, integrals are crucial. Here, you integrate the function over the interval and then divide by the interval's length.

Unlike differentiation, which finds the rate of change, integration aggregates values to find totals. In this exercise, the function given was \( f(x) = 1 + \cos(x) \). To find the average value of this function over the interval \([0, 2\pi]\), you find the total area under the curve and then divide by the total interval length. This method provides a tangible application of integral calculus, showcasing its practical utility.

Trigonometric Functions

Trigonometric functions are vital in many areas of mathematics, including integral calculus. The function \( f(x) = 1 + \cos(x) \) includes the cosine function, a fundamental trigonometric function noted for its wave-like patterns and periodicity.

Cosine is an even function, meaning it is symmetric about the y-axis. For integral calculations, this symmetry often simplifies the process, especially over intervals that mirror this symmetry, like \([0, 2\pi]\). The integral of \( \cos(x) \) over \([0, 2\pi]\) results in zero because the positive and negative areas cancel each other out. Understanding these properties aids in solving problems quickly and accurately.

• Wave patterns repeat every \(2\pi\) for cosine
• Cosine has a range from -1 to 1
• Key points: peak (1), trough (-1), and zero crossings (0)

Being familiar with trigonometric functions and their integrals is essential for dealing with challenges involving periodic functions.

Definite Integral

A definite integral yields a number representing the signed area under a curve from one point to another on the x-axis. The 'signed' aspect means areas above the x-axis are positive and areas below are negative. This characteristic is fundamentally different from an indefinite integral, which results in a family of functions.

In the exercise, the definite integral \( \int_{0}^{2\pi} (1 + \cos(x)) \; dx \) was used. It precisely calculated the total 'net' area of the curve from \( x = 0 \) to \( x = 2\pi \). After integrating, the areas which were symmetric above and below the x-axis canceled out.

• Calculates total net area
• Requires limits of integration: lower \(a\) and upper \(b\)
• Yields a specific numerical value

Understanding definite integrals is vital in finding quantities such as accumulated change, area, and averages. It solidifies the connection between algebraic and geometric interpretations of calculus.