Problem 1 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The average value of \( f(x) = \cos(x) \) on \([0, \pi/2]\) is \( \frac{2}{\pi} \).

Step 1: Understand the Average Value Formula

To find the average value of a continuous function \( f(x) \) on a closed interval \( [a, b] \), we use the formula: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \]. Here, \( f(x) = \cos(x) \) and \( I = [0, \pi/2] \).

Step 2: Identify the Interval and Setup the Integral

Identify \( a = 0 \) and \( b = \pi / 2 \). Next, set up the integral for the average value formula: \[ \text{Average value} = \frac{1}{\pi/2 - 0} \int_0^{\pi/2} \cos(x) \, dx \].

Step 3: Simplify the Coefficient

Simplify the coefficient in front of the integral: \( \frac{1}{\pi/2} = \frac{2}{\pi} \). So the expression becomes: \[ \frac{2}{\pi} \int_0^{\pi/2} \cos(x) \, dx \].

Step 4: Calculate the Integral

Evaluate the integral \( \int_0^{\pi/2} \cos(x) \, dx \). The antiderivative of \( \cos(x) \) is \( \sin(x) \). Compute it over the interval \([0, \pi/2]\): \[ \left[ \sin(x) \right]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1 \].

Step 5: Compute the Average Value

Multiply the result of the integral by the coefficient: \( \frac{2}{\pi} \times 1 = \frac{2}{\pi} \). Therefore, the average value of \( f(x) = \cos(x) \) over the interval \([0, \pi/2]\) is \( \frac{2}{\pi} \).

Key Concepts

Integral Calculus

Integral calculus is a vast field of mathematics that deals with integrals, which are essentially the 'reverse' of derivatives. It helps us to understand the accumulation of quantities, such as areas under curves or the total quantity given a rate of change.

Integrals are used extensively in fields like physics, engineering, and economics to solve various real-world problems. A key part of integral calculus is identifying how continuously changing equations can result in a finite area or quantity. This is where the concept of integrating a function over an interval comes into play, as seen in calculating the average value of a function.

In our context, we used integral calculus to find the average value of the function \( f(x) = \cos(x) \) over the interval \([0, \pi/2]\). By simplifying and evaluating the integral, we can determine the mean value of the cosine function over this specified range.

Definite Integral

The definite integral is a specific type of integral with upper and lower limits, which allows us to calculate the net area under a curve between two points. In the formula for the average value of a function, the definite integral is pivotal.

Mathematically, for a function \( f(x) \) over an interval \([a, b]\), the definite integral is represented as \( \int_a^b f(x) \, dx \).

• It provides a single numeric value representing the total accumulation of \( f(x) \) from \( a \) to \( b \).
• The area can be positive or negative based on the position of the function relative to the x-axis.

The calculation of this integral essentially gives us the sum of infinite small areas formed under the curve \( f(x) \), resulting in the net area caught within the bounds. This way, the definite integral helps us determine the average value by integrating the function \( \cos(x) \) over \([0, \pi/2]\).

Trigonometric Functions

Trigonometric functions, like cosine and sine, are fundamental in mathematics, particularly when constructing and analyzing periodic phenomena. These functions are deeply entrenched in geometry and have wide applications.

Their periodic nature makes them essential when dealing with cycles and oscillations, as they encapsulate rotations, waves, and harmonics.

In the given exercise, the function \( f(x) = \cos(x) \) is part of this family, representing these cycles. Trigonometric functions are integral in solving not only problems in calculus but also in algebra and geometry, as they help bridge gaps between linear and angular relationships. Understanding these functions allows us to interpret various patterns and phenomena in mathematics and the natural world.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function. It answers the reverse question posed by differentiation: if \( F(x) \) is the antiderivative of \( f(x) \), then \( F'(x) = f(x) \). This concept is critical when solving integrals.

For the function \( \cos(x) \), its antiderivative is \( \sin(x) \). This relationship was essential in evaluating the definite integral \( \int_0^{\pi/2} \cos(x) \, dx \), as it allowed us to directly compute the value by substituting the upper and lower limits into \( \sin(x) \).

• Antiderivatives offer a way to "undo" differentiation.
• Each function can have many antiderivatives; they differ by a constant.

The power of antiderivatives shines particularly in solving area problems or when reconstructing original quantities from their rates of change.