Question

In Exercises \(31 - 36 ,\) find the average value of the function on the interval, using antiderivatives to compute the integral. $$y = \sec x \tan x , \quad \left[ 0 , \frac { \pi } { 3 } \right]$$

Short answer

The average value of the function \(sec(x)tan(x) \) over the interval \([0, \pi/3]\) is \(\frac{3}{\pi} \cdot ln|2 + \sqrt{3}|\)

Step-by-step solution

Set up The Formula

Set up the formula for the average value of the function, which is given by \(AvgValue(f) = \frac{1}{b-a}\int_{a}^{b}f(x)dx, \) In this case, \(a = 0, b = \pi/3, \) and \(f(x) = sec(x)tan(x). \) Therefore, the average value is given by \(AvgValue(f) = \frac{1}{\pi/3-0}\int_{0}^{\pi/3}sec(x)tan(x)dx.\)

Compute The Integral

The definite integral of the function \(sec(x)tan(x) \) from \(0\) to \(\pi/3\) is the antiderivative of \(secx \) which is \(ln|secx + tanx|\), evaluated from \(0\) to \(\pi/3\). That will yield \(\int_{0}^{\pi/3}sec(x)tan(x)dx= ln|sec(\pi/3) + tan(\pi/3)| - ln|sec(0) + tan(0)| = ln|2 + \sqrt{3}| - ln|1 + 0| = ln|2 + \sqrt{3}|\)

Compute The Average Value

The average value is computed by substituting the value of the integral obtained in step 2 into the formula obtained in step 1. So we have \(AvgValue(f) =\frac{1}{\pi/3-0} \cdot ln|2 + \sqrt{3}| = \frac{3}{\pi} \cdot ln|2 + \sqrt{3}|\)

Key concepts

Definite Integral

Understanding the concept of a definite integral is crucial when you're trying to find the average value of a function over an interval. A definite integral is essentially the area under the curve of the function between two points along the x-axis. In the context of the given problem, we’re looking at the area under the curve of the function f(x) = sec(x)tan(x) from x = 0 to x = \frac{\boldsymbol{\text{\textpi}}}{3}.

The notation for the definite integral of a function f(x) from a to b is written as \[ \int_{a}^{b} f(x) \, dx \. \] This formula would be used to determine the bounded area, which can sometimes represent physical quantities like displacement or the amount of fluid that fills a particular shape.

When calculating the definite integral, certain properties can make this process easier, such as the Power Rule, and for more complex functions, integral tables or computer algebra systems might be necessary.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the inverse operation of differentiation. If a function f(x) is the derivative of some function F(x), then F(x) is an antiderivative of f(x). When you're working out a definite integral, as we need to in the addressed exercise, you're finding the antiderivative of the function and then evaluating it at the upper and lower bounds of the interval.

The antiderivative of sec(x)tan(x) in the given problem is \[ \ln|\sec(x) + \tan(x)| \.\] This result is then evaluated over the interval [0, \frac{\boldsymbol{\text{\textpi}}}{3}] to find the integral’s value, which is needed to compute the average value of the function. Antiderivatives can often be found using techniques such as substitution, integration by parts, or by using standard formulas.

Secant Tangent Identity

In the calculation of the average value of a function using antiderivatives, particularly with trigonometric functions, it's beneficial to understand the identities that simplify the integration process. One such identity is the secant tangent identity, which states that

\[ \sec x \cdot \tan x = \frac{\sin x}{\cos^{2} x} \.\]

This identity comes from the definitions of secant and tangent in terms of sine and cosine, where sec(x) is 1/cos(x), and tan(x) is sin(x)/cos(x). However, in our specific case, we do not need to use this identity directly, since the integral of sec(x)tan(x) is well-known and leads directly to the natural logarithm function, as used in the solution.

Integral Properties

When dealing with integrals, there are important properties to remember that can simplify the process, such as the linearity of integration and the fundamental theorem of calculus. Linearity allows us to break down complex integrals into simpler parts, while the fundamental theorem of calculus connects derivatives with integrals and provides a strategy for evaluating definite integrals.

In our example, the property that \[ \int \ln|u| \, dx = u \ln|u| - u + C \.\] is used. Here, C is the constant of integration and is not needed in the definite integral.

Integration properties are essential for calculating definite integrals accurately, and in the case of our exercise, we have to apply these properties systematically to find the average value of the function over the given interval. Remember, even the most complex integral can usually be approached by breaking it down using these fundamental properties.