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Magazine Chapter 4: Problem 12
Find the average value of the function on the given interval. $$ g(x)=\tan x \sec ^{2} x ; \quad[0, \pi / 4] $$
Short Answer
Expert verified
The average value is \( \frac{2}{\pi} \).
Step by step solution
01
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 \( \frac{4}{\pi} \int_{0}^{\pi/4} \tan x \sec^2 x \, dx \).
02
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 \( du = \sec^2 x \, dx \).
03
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 \( \int_{0}^{1} u \, du \).
04
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 \( \left. \frac{u^2}{2} \right|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \).
05
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 \( \frac{2}{\pi} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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鈥檚 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.
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.
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鈥檚 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.
