Problem 12 - Calculus

Question

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

Step-by-Step Solution

Answer

The average value is \( \frac{2}{\pi} \).

Step 1: Understand the Problem

To find the average value of a function \( g(x) \) on the interval \([a, b]\), we use the formula for the average value, which is \( \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \). Here, \( g(x) = \tan x \sec^2 x \), and the interval is \([0, \pi/4]\). So, we need to compute \( \frac{4}{\pi} \int_{0}^{\pi/4} \tan x \sec^2 x \, dx \).

Step 2: Simplify the Integral Expression

Notice that the integral \( \int \tan x \sec^2 x \, dx \) is suitable for a substitution. Recall that \( \sec^2 x = (1+\tan^2 x) \) and that the derivative of \( \tan x \) is \( \sec^2 x \). This suggests using the substitution \( u = \tan x \), so \( du = \sec^2 x \, dx \).

Step 3: Change of Variables

Substituting \( u = \tan x \), the differential \( du = \sec^2 x \, dx \). Thus, the integral becomes \( \int u \, du \). The limits of integration change accordingly: when \( x = 0 \), \( u = \tan 0 = 0 \), and when \( x = \pi/4 \), \( u = \tan(\pi/4) = 1 \). So, the new integral is \( \int_{0}^{1} u \, du \).

Step 4: Solve the Integral

Evaluate the integral \( \int_{0}^{1} u \, du \) using the power rule for integration: \( \int u \, du = \frac{u^2}{2} \). Therefore, compute \( \left. \frac{u^2}{2} \right|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \).

Step 5: Calculate the Average Value

The average value is \( \frac{4}{\pi} \times \frac{1}{2} = \frac{2}{\pi} \). Hence, the average value of the function \( g(x) = \tan x \sec^2 x \) over the interval \([0, \pi/4]\) is \( \frac{2}{\pi} \).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify integration by changing the variable of integration. It is particularly useful when dealing with composite functions. Here’s how it works:

• You choose a substitution, usually denoted as \( u \), that simplifies the integral. Typically, \( u \) is a function inside the integral whose derivative also appears in the integral.
• Once \( u \) is chosen, compute \( du \), which is the derivative of \( u \) with respect to \( x \), multiplied by \( dx \).
• Substitute both \( u \) and \( du \) into the integral, changing the limits of integration if it is a definite integral.
• Integrate with respect to \( u \), then convert back to the original variable if needed.

In our problem, the substitution \( u = \tan x \) was chosen because its derivative, \( \sec^2 x \), is present in the integral. This made the integral \( \int \tan x \sec^2 x \, dx \) transform into \( \int u \, du \), which is straightforward to integrate.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, play a crucial role in many areas of mathematics. They are especially important in studying angles and periodic functions. Here are a few key points about the trigonometric functions involved in this exercise:

• The tangent function, \( \tan x \), is the ratio of the opposite side to the adjacent side in a right-angled triangle. In terms of sine and cosine, \( \tan x = \frac{\sin x}{\cos x} \).
• The secant function, \( \sec x \), is the reciprocal of the cosine function, i.e., \( \sec x = \frac{1}{\cos x} \). Thus, \( \sec^2 x = \left(\frac{1}{\cos x}\right)^2 \).
• These functions exhibit periodic behavior and have important derivative relationships. Notably, the derivative of \( \tan x \) is \( \sec^2 x \), which is why this substitution worked well in the integration problem.

Understanding these relationships helps in solving integrals and trigonometric equations effectively.

Definite Integral

A definite integral is used to calculate the accumulation of quantities, such as areas under curves. In the given exercise, we needed to compute a definite integral to find the average value of a function over an interval. Here’s what you should know:

• The definite integral from \( a \) to \( b \) of a function \( f(x) \), denoted as \( \int_{a}^{b} f(x) \, dx \), represents the net area between the function and the x-axis from \( x = a \) to \( x = b \).
• To evaluate, you first find an antiderivative \( F(x) \) of \( f(x) \), and then compute \( F(b) - F(a) \).
• When calculating the average value of a function over an interval \([a, b]\), you adjust the definite integral by \( \frac{1}{b-a} \), normalizing it over the interval's length.

In the solution, the integral was \( \int_{0}^{\pi/4} \tan x \sec^2 x \, dx \), which required substitution to solve efficiently. This integral calculated the area under the function from 0 to \( \frac{\pi}{4} \), leading us to find the average value as \( \frac{2}{\pi} \).