Question

Find the average value of the function \(f(x)=\frac{1}{x^{2}+1}\) over the interval \([-3,3]\).

Short answer

The average value is approximately 0.416.

Step-by-step solution

Write the Formula for the Average Value

The formula for the average value of a continuous function \( f(x) \) over the interval \([a, b]\) is given by \( \frac{1}{b-a} \int_a^b f(x) \, dx \). Here, \( a = -3 \) and \( b = 3 \).

Set Up the Integral

We need to evaluate the integral of \( f(x) = \frac{1}{x^2+1} \) from \(-3\) to \(3\). The integral is: \( \int_{-3}^{3} \frac{1}{x^2+1} \, dx \).

Recognize the Integral as a Standard Form

The integral \( \int \frac{1}{x^2+1} \, dx \) is a standard integral and its antiderivative is the arctangent function: \( \tan^{-1}(x) \).

Evaluate the Definite Integral

Use the antiderivative to evaluate the definite integral: \( \left[ \tan^{-1}(x) \right]_{-3}^{3} = \tan^{-1}(3) - \tan^{-1}(-3) \).

Calculate the Arctangents

Recognizing that \( \tan^{-1}(-x) = -\tan^{-1}(x) \), the expression simplifies to \( \tan^{-1}(3) + \tan^{-1}(3) = 2\tan^{-1}(3) \).

Compute the Average Value

Apply the average value formula: \( \frac{1}{3 - (-3)} \times 2\tan^{-1}(3) = \frac{1}{6} \times 2\tan^{-1}(3) = \frac{1}{3} \tan^{-1}(3) \).

Final Calculation

Using a calculator, \( \tan^{-1}(3) \approx 1.249\), so the average value is \( \frac{1}{3} \times 1.249 \approx 0.416 \).

Key concepts

Definite Integral

To find the average value of a function over a given interval, we use the concept of a definite integral. The definite integral helps us to compute the accumulated area under a curve defined by the function over a specific interval.
The process of finding the average value of a function involves:

• Identifying the function over the specified interval \([a, b]\).
• Calculating the integral of the function from \(a\) to \(b\), denoted as \( \int_a^b f(x) \, dx \).
• Dividing the integral result by the length of the interval \(b - a\).

By performing this step, we can determine how the function behaves on average over that interval. This technique does not just tell us about specific values of the function but instead helps us understand its overall trend or central tendency within that range.

Antiderivative

An antiderivative is essentially the opposite operation of taking a derivative. While a derivative gives us the rate of change or slope of a function at any point, the antiderivative tells us which function could have resulted in that rate of change.
In the problem, to find the definite integral of \( f(x) = \frac{1}{x^2+1} \), we use the antiderivative.

• The antiderivative of \( \frac{1}{x^2+1} \) is the arctangent function, denoted \( \tan^{-1}(x) \).
• This means that if we take the derivative of \( \tan^{-1}(x) \), we get back \( \frac{1}{x^2+1} \).

Understanding antiderivatives is crucial for solving integrals, especially when determining the average value of a function. The evaluated antiderivative at the bounds of the interval gives the net area or the accumulated value, which is then averaged over the interval.

Arctangent Function

The arctangent function, often represented as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function. It helps us solve problems where we need to find the angle whose tangent is a given number.
In calculus, the arctangent function is important as an antiderivative for integrals involving \( \frac{1}{x^2+1} \).

• The evaluation \( \left[ \tan^{-1}(x) \right]_{-3}^{3} \) provides us with \( \tan^{-1}(3) - \tan^{-1}(-3) \).
• Since \( \tan^{-1}(-x) = -\tan^{-1}(x) \), this simplifies to \( 2\tan^{-1}(3) \).
• For numerical calculations, \( \tan^{-1}(3) \approx 1.249 \).

Mastering the arctangent function allows us to handle integrals of this type efficiently, making it a fundamental part of solving calculus problems that involve average values of functions over specific intervals.