Problem 126 - Calculus Volume 1

Question

Show that the average value of \(\sin ^{2} t\) over \([0,2 \pi]\) is equal to 1\(/ 2\) Without further calculation, determine whether the average value of \(\sin ^{2} t\) over \([0, \pi]\) is also equal to 1\(/\)2.

Step-by-Step Solution

Answer

The average value of \( \sin^2(t) \) over both intervals is \(1/2\).

Step 1: Understand the Average Value Formula

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:\[\frac{1}{b-a} \int_a^b f(x) \, dx\]For this exercise, the function is \( f(t) = \sin^2(t) \) and the interval is \([0, 2\pi]\).

Step 2: Simplify Using Trigonometric Identity

To simplify the integral, use the trigonometric identity for \( \sin^2(t) \) which is:\[\sin^2(t) = \frac{1 - \cos(2t)}{2}\]This helps in simplifying the integration.

Step 3: Set Up the Integral

Substitute \( \sin^2(t) \) in the average formula with its identity:\[\frac{1}{2\pi} \int_0^{2\pi} \frac{1 - \cos(2t)}{2} \, dt\]This is the integral we need to solve next.

Step 4: Split and Evaluate the Integral

Split the integral into two:\[\frac{1}{4\pi} \int_0^{2\pi} 1 \, dt - \frac{1}{4\pi} \int_0^{2\pi} \cos(2t) \, dt\]Calculate each part separately. The first part results in \(\frac{1}{4\pi} imes (2\pi) = \frac{1}{2}\).The second part, due to periodicity and being a complete cycle of cosine, evaluates to 0.

Step 5: Calculate the Result

Combining both results from Step 4 gives:\[\frac{1}{2} - 0 = \frac{1}{2}\]Thus, the average value of \( \sin^2(t) \) over \([0, 2\pi]\) is \(\frac{1}{2}\).

Step 6: Analyze the Other Interval Without Calculation

The function \( \sin^2(t) \) is symmetric about \(\pi\) and its periodic nature implies that its integral over each complete half-period \([0, \pi]\) is equal to the other half \([\pi, 2\pi]\). Thus, the average value over \([0, \pi]\), without any further computation, is the same as over \([0, 2\pi]\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that deals with finding the accumulation of quantities. It plays a pivotal role in problems involving area under curves, volumes, central points, and much more.
In the context of average values, integral calculus helps us determine the mean of a continuous function over a specified interval. This is represented by the formula:

• \( \frac{1}{b-a} \int_a^b f(x) \, dx \)

This formula essentially gives you the average height of the function \( f(x) \) over the range from \( a \) to \( b \).
For the problem at hand, we used integral calculus to find the average value of \( \sin^2(t) \), which involves setting up the integral of this function over the interval \([0, 2\pi]\). Integrating the function by substituting with the trigonometric identity simplifies the process and ensures more straightforward calculations.

Each time we solve for an average using integral calculus, we are essentially capturing the essence of the function's behavior over an interval, smoothing out moments of extreme highs and lows.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variables involved. These identities are incredibly useful for simplifying complex trigonometric equations and integrals.
In this exercise, we encountered the identity for \( \sin^2(t) \):

• \( \sin^2(t) = \frac{1 - \cos(2t)}{2} \)

This notable identity arises from the double angle identities for cosine, and it is used to reformulate the square of the sine function into a form that can be integrated more easily.

The ability to recognize and apply such identities is crucial for efficiently solving integrals, especially when dealing with squares of trigonometric functions or more complicated expressions. Knowing these identities allows us to transform and tackle integrals that might otherwise seem daunting, leading to simpler computations and clearer solutions.

Periodic Functions

Periodic functions are functions that repeat their values at regular intervals; their graphs exhibit symmetry. One of the most common examples of periodic functions are the trigonometric functions, such as sine and cosine.
The function \( \sin^2(t) \) is periodic with a period of \( \pi \), meaning it repeats its values every \( \pi \) units. This property is significant when considering the integral of trigonometric functions over their periods.

This exercise leverages the periodic nature of \( \sin^2(t) \), particularly since the interval \([0, 2\pi]\) constitutes precisely two full cycles. This periodicity allows us to evaluate the integral by understanding that the contribution from each half-period is equal, simplifying the computation of the average.
Because of this periodic symmetry, the average value calculated over a full cycle, \([0, 2\pi]\), conveniently applies to \([0, \pi]\) as well. Recognizing these symmetrical properties can drastically reduce computational efforts and provide quick insights without detailed arithmetic.