Question

Find the average value of the function on the given interval. $$ g(x)=\tan x \sec ^{2} x ; \quad[0, \pi / 4] $$

Short answer

Average value is \(\frac{2}{\pi}\).

Step-by-step solution

Understand the Formula for Average Value

To find the average value of a function over an interval \([a, b]\), we use the formula: \(\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). For this problem, the function is \(g(x) = \tan x \sec^2 x\) and the interval is \[0, \frac{\pi}{4}\].

Perform Integration Setup

Substitute \(a = 0\), \(b = \frac{\pi}{4}\), and \(f(x) = \tan x \sec^2 x\) into the average value formula. The expression becomes: \\[\frac{1}{\frac{\pi}{4} - 0} \int_{0}^{\frac{\pi}{4}} \tan x \sec^2 x \, dx.\]

Simplify the Expression

Simplify \(\frac{1}{\frac{\pi}{4}} = \frac{4}{\pi}\). So the expression is: \\[\frac{4}{\pi} \int_{0}^{\frac{\pi}{4}} \tan x \sec^2 x \, dx.\]

Evaluate the Integral

Use substitution to evaluate the integral. Let \(u = \tan x\), so \(du = \sec^2 x \, dx\). The integral becomes: \\[\int u \, du = \frac{u^2}{2}.\]Substitute back \(u = \tan x\) to get: \\[\frac{\tan^2 x}{2}\] evaluated from \(0\) to \(\frac{\pi}{4}.\) At \(x = \frac{\pi}{4}\), \ \tan x = 1\.At \(x = 0\), \ \tan x = 0\.

Substitute Integration Results

Evaluate the result: \\[\frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.\]Substitute back into the average value formula: \\[\frac{4}{\pi} \times \frac{1}{2} = \frac{2}{\pi}.\]

State the Final Result

The average value of the function \(g(x) = \tan x \sec^2 x\) over the interval \([0, \frac{\pi}{4}]\) is \(\frac{2}{\pi}.\)

Key concepts

Integration Techniques

When dealing with the integration of functions, especially in calculus, it is important to use the right techniques to make the process easier and more straightforward. In this exercise, we use integration to find the average value of a function over a specific interval. To calculate this, we apply the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] In the problem at hand, we're dealing with the function \( g(x) = \tan x \sec^2 x \) over the interval \([0, \frac{\pi}{4}]\). **Using Substitution** - We often use substitution when faced with complex integrals. It simplifies the function inside the integral, turning it into something easier to work with.- For this specific function, setting \( u = \tan x \) simplifies the integral. The original integral becomes \( \int \tan x \sec^2 x \, dx \), which through substitution transforms to \( \int u \, du \). By recognizing the derivative \( du = \sec^2 x \, dx \), we make the integral much more manageable. In summary, the technique of substitution can make seemingly complex integrals approachable by converting them into simpler ones.

Trigonometric Functions

Trigonometric functions like tangent and secant appear frequently in calculus, especially in exercises involving integration. Understanding their properties can greatly assist in calculations and simplifications. **Understanding Tangent and Secant**- **Tangent (\( \tan x \))**: Represents the ratio of the opposite side to the adjacent side in a right triangle of angle \( x \).- **Secant (\( \sec x \))**: This is the reciprocal of cosine, \( \sec x = \frac{1}{\cos x} \).In our problem, the function \( g(x) = \tan x \sec^2 x \) integrates these two trigonometric functions. Knowing that \( \sec^2 x \) is the derivative of \( \tan x \) makes it easier to perform integration by substitution, as seen in the previous section.When integrating trigonometric functions, always consider: - Their derivatives (e.g., \( \, \frac{d}{dx} (\tan x) = \sec^2 x \)).- Their identities (e.g., \( \sec x = \frac{1}{\cos x} \)).Using these simple properties, you can transform and simplify complex expressions involving these trigonometric functions.

Calculus

Calculus deals with the study of change and motion. At its core, it involves the concepts of integrals and derivatives, which are foundational in solving real-world problems involving rates of change.**Using Calculus to Find Average Value**- Calculus enables us to find the average value of functions over certain intervals using integration.In this exercise, we calculated the average value of a function \( g(x) \) over the interval \([0, \frac{\pi}{4}]\). - The process involves integrating the function over its interval and then multiplying by the reciprocal of the length of the interval The result we obtained was \( \frac{2}{\pi} \), which signifies the average value of the function across the given interval. **Importance of Understanding**Calculus is crucial for:

• Solving problems in physics and engineering where rates of change are necessary.
• Calculating areas and volumes using integration.
• Understanding the behavior of functions and their rates of increase or decrease.

By mastering calculus concepts, you'll be able to understand and solve a wide range of mathematical problems much more efficiently.