Problem 48 - Calculus Single Variable

Question

In each of Exercises \(43-52\) calculate the average of the given expression over the given interval. $$ \sin ^{2}(2 x) \quad 0 \leq x \leq \pi / 2 $$

Step-by-Step Solution

Answer

The average value of \( \sin^2(2x) \) over \( [0, \frac{\pi}{2}] \) is \( \frac{1}{2} \).

Step 1: Understand the Problem

We are asked to calculate the average value of the function \( \sin^2(2x) \) over the interval \( [0, \frac{\pi}{2}] \). The average value of a function \( f(x) \) over an interval \([a, b]\) is given by \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \).

Step 2: Apply the Average Value Formula

Apply the formula for the average value of a function: \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} \sin^2(2x) \, dx \] for the interval \([0, \frac{\pi}{2}]\). Here, \(a = 0\) and \(b = \frac{\pi}{2}\).

Step 3: Simplify the Integral

Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \), the integral becomes: \[\int_{0}^{\frac{\pi}{2}} \sin^2(2x) \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(4x)}{2} \, dx.\]

Step 4: Evaluate the Integral

Separate the integral: \[\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, dx - \int_{0}^{\frac{\pi}{2}} \frac{\cos(4x)}{2} \, dx.\]The first integral: \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, dx = \frac{1}{2} \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}. \)For the second integral, use substitution: Let \( u = 4x \), then \( du = 4 \, dx \) or \( dx = \frac{1}{4} du \).This gives \(\int \frac{\cos(4x)}{2} \, dx = \frac{1}{8} \int \cos(u) \, du = \frac{1}{8} \sin(u) + C.\)Therefore the evaluated definite integral: \[\int_{0}^{\frac{\pi}{2}} \frac{\cos(4x)}{2} \, dx = \left[ \frac{1}{8} \sin(4x) \right]_{0}^{\frac{\pi}{2}} = \left( \frac{1}{8} \cdot \sin(2\pi) \right) - \left( \frac{1}{8} \cdot \sin(0) \right) = 0 \]

Step 5: Calculate the Average Value

Now substitute the evaluated integrals back into the average formula: \[ \text{Average} = \frac{1}{\frac{\pi}{2} - 0} \left( \frac{\pi}{4} - 0 \right) = \frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}. \]

Key Concepts

Average Value of a Function

The average value of a function is a fundamental concept in calculus used to find a single value that represents the overall behavior of a function over a given interval. In our exercise, we aim to find the average value of the function \( \sin^2(2x) \) over the interval \([0, \frac{\pi}{2}]\). The formula for finding the average value of a continuous function \( f(x) \) over the interval \([a, b]\) is:\[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \].
Using this formula helps in summarizing the entire area under the curve of \( f(x) \) over \([a, b]\) into a singular average height, though it does not necessarily match any specific function value. This concept is crucial, especially in applications dealing with physics and engineering, as it often simplifies complex problems into understandable numbers.
In our example, we used \( a = 0 \) and \( b = \frac{\pi}{2} \), resulting in the average value being \( \frac{1}{2} \). Hence, this gives us an approximate general value for \( \sin^2(2x) \) on the specified interval.

Integration

Integration is a key tool in calculus that helps in calculating the area under a curve, among many other applications. In this exercise, integration was utilized to calculate the average value of \( \sin^2(2x) \) over \([0, \frac{\pi}{2}]\). For such tasks, the definite integral is the appropriate choice because it calculates not just an antiderivative, but the net area within a certain range.
The integral of the given trigonometric function was tackled by employing a known trigonometric identity, making it easier to evaluate. After using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \), the integral simplifies to calculations of simpler functions. Splitting the problem further into integral parts like \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, dx \) and \( \int_{0}^{\frac{\pi}{2}} \frac{\cos(4x)}{2} \, dx \) allows us to apply integration techniques straightforwardly.
This decomposition highlights how integration helps in clear step-by-step gradual solution building, particularly in handling complex expressions.

Trigonometric Identities

Trigonometric identities are powerful tools in calculus and beyond, providing methods to simplify complex trigonometric functions. For the function \( \sin^2(2x) \), this exercise made use of one such identity: \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \).
By employing this identity, the integral transformed from an inconvenient square of a sine function into a more manageable combination involving cosine. This transformation not only allowed easier integration but also avoided potential pitfalls in recognition and computation of basic integrals. Typically, recognizing and applying the correct identities can significantly ease the computational process in trigonometric integrations.

• Reduces complexity of functions.
• Enables easier integration or differentiation.
• Widely applicable in many areas, including physics and engineering.

Understanding and leveraging these identities can streamline the approach to many calculus problems, removing sometimes burdensome algebraic steps.