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Magazine Chapter 7: Problem 66
Calculate the average value of \( f(x) = x \sec^2 x \) on the interval \( [0, \frac{\pi}{4}] \).
Short Answer
Expert verified
The average value is \(1 - \frac{2 \ln 2}{\pi}\).
Step by step solution
01
Understand the Formula for Average Value
The average value of a function \( f(x) \) on a closed interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). In this case, the function is \( f(x) = x \sec^2 x \) and the interval is \( [0, \frac{\pi}{4}] \).
02
Set Up the Integral
Substitute \( f(x) = x \sec^2 x \) and the interval into the formula for average value: \[ \frac{1}{\frac{\pi}{4} - 0} \int_{0}^{\frac{\pi}{4}} x \sec^2 x \ dx \]. This simplifies to: \[ \frac{4}{\pi} \int_{0}^{\frac{\pi}{4}} x \sec^2 x \ dx \].
03
Solve the Integral Using Integration by Parts
Use integration by parts which states \( \int u \, dv = uv - \int v \, du \). Let \( u = x \) and \( dv = \sec^2 x \, dx \), then \( du = dx \) and \( v = \tan x \). Apply the formula:\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \].
04
Evaluate the Integral of \( \tan x \)
The integral of \( \tan x \) is \(-\ln |\cos x| + C\). Thus,\[ \int x \sec^2 x \, dx = x \tan x + \ln |\cos x| + C \].
05
Calculate the Definite Integral
Evaluate \[ x \tan x + \ln |\cos x| \] from \(0\) to \(\frac{\pi}{4}\):\[ \left[ x \tan x + \ln |\cos x| \right]_{0}^{\frac{\pi}{4}} = \left( \frac{\pi}{4} \cdot 1 + \ln \cos \frac{\pi}{4} \right) - \left( 0 + \ln \cos 0 \right) \].
06
Simplify the Expression
\( \tan \frac{\pi}{4} = 1 \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( \cos 0 = 1 \). Substitute these into the expression:\[ \frac{\pi}{4} + \ln \frac{1}{\sqrt{2}} - 0 - 0 = \frac{\pi}{4} - \frac{1}{2} \ln 2 \].
07
Calculate the Average Value
Multiply the result by \(\frac{4}{\pi}\):\[ \frac{4}{\pi} \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right) = 1 - \frac{2 \ln 2}{\pi} \]. This is the average value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a valuable technique for solving integrals, especially when the integrand is a product of two functions. The technique is based on the formula \( \int u \, dv = uv - \int v \, du \), which can be thought of as a way to reverse the product rule for differentiation. To use it effectively, one needs to properly identify and choose the parts of the integral to assign as \( u \) and \( dv \).
Here's the key to picking \( u \) and \( dv \): generally, choose \( u \) to be a function that becomes simpler when differentiated, and \( dv \) to be a function that is easily integrated. In our example, we set \( u = x \) (since the derivative, \( dx \), is simpler) and \( dv = \sec^2 x \, dx \).
Here's the key to picking \( u \) and \( dv \): generally, choose \( u \) to be a function that becomes simpler when differentiated, and \( dv \) to be a function that is easily integrated. In our example, we set \( u = x \) (since the derivative, \( dx \), is simpler) and \( dv = \sec^2 x \, dx \).
- Differentiate \( u \): \( du = dx \).
- Integrate \( dv \): \( v = \tan x \).
Definite Integral
A definite integral is a type of integral with upper and lower bounds, represented as \( \int_{a}^{b} f(x) \, dx \). It calculates the net area under the curve defined by \( f(x) \) from \( x = a \) to \( x = b \). This "net area" takes into account areas above the x-axis as positive and those below as negative.
In the problem, after using integration by parts, we evaluated the definite integral \( \int_{0}^{\frac{\pi}{4}} x \sec^2 x \, dx \). The process involved calculating the anti-derivative at the upper and lower limits, then finding the difference. The fundamental theorem of calculus tells us that the definite integral can be found as \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \).
In the problem, after using integration by parts, we evaluated the definite integral \( \int_{0}^{\frac{\pi}{4}} x \sec^2 x \, dx \). The process involved calculating the anti-derivative at the upper and lower limits, then finding the difference. The fundamental theorem of calculus tells us that the definite integral can be found as \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \).
- Upper limit: \( x \tan x + \ln |\cos x| \) evaluated at \( \frac{\pi}{4} \).
- Lower limit: the same function evaluated at \( 0 \).
Trigonometric Functions
Trigonometric functions are fundamental to calculus, and they frequently appear in integration problems. In this exercise, we primarily dealt with \( \sec^2 x \), \( \tan x \), and \( \cos x \). Understanding their properties and relationships is crucial to solving integrals that involve them.
The derivative of the tangent function is \( \sec^2 x \), which provided a convenient choice for \( dv \) when applying integration by parts. Similarly, recognizing that the integral of \( \tan x \) is \( -\ln |\cos x| + C \) helped us solve part of the definite integral.
The derivative of the tangent function is \( \sec^2 x \), which provided a convenient choice for \( dv \) when applying integration by parts. Similarly, recognizing that the integral of \( \tan x \) is \( -\ln |\cos x| + C \) helped us solve part of the definite integral.
- \( \tan \frac{\pi}{4} = 1 \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), values used in simplifying the result.
- These trigonometric values are useful benchmarks when working with common angles.
