Question

Find the average value of the function on the given interval. $$ f(x)=\frac{x}{\sqrt{x^{2}+16}} ; \quad[0,3] $$

Short answer

The average value of the function on the interval [0, 3] is \( \frac{1}{3} \).

Step-by-step solution

Understand the formula for average value

The average value of a continuous function \( f(x) \) on an interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_a^b f(x) \, dx \). Thus, for this problem, the average value formula becomes \( \frac{1}{3-0} \int_0^3 \frac{x}{\sqrt{x^2+16}} \, dx \).

Set up the integral

We need to evaluate the integral \( \int_0^3 \frac{x}{\sqrt{x^2+16}} \, dx \). This integral can be solved using the substitution method.

Choose a suitable substitution

Let \( u = x^2 + 16 \), then \( du = 2x \, dx \), and \( \frac{du}{2} = x \, dx \). The limits of integration change accordingly: when \( x = 0 \), \( u = 16 \); when \( x = 3 \), \( u = 3^2 + 16 = 25 \).

Rewrite the integral in terms of \( u \)

The integral becomes \( \frac{1}{2} \int_{16}^{25} \frac{1}{\sqrt{u}} \, du \).

Integrate with respect to \( u \)

The integral \( \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \) simplifies to \( \frac{1}{2} [2\sqrt{u}] + C = \sqrt{u} + C \). Evaluate \( \sqrt{u} \) from \( 16 \) to \( 25 \).

Evaluate the definite integral

Substituting the limits, \( \sqrt{25} - \sqrt{16} = 5 - 4 = 1 \).

Calculate the average value

Substitute the result of the integral back into the average value formula: \( \frac{1}{3 - 0} \times 1 = \frac{1}{3} \).

Conclude

The average value of the function \( f(x) = \frac{x}{\sqrt{x^2+16}} \) on the interval \([0, 3]\) is \( \frac{1}{3} \).

Key concepts

Definite Integral

A definite integral represents the signed area under a curve of a given function within a specified interval. It is calculated using the lower and upper limits of the interval, denoted as \( \int_a^b f(x) \, dx \). The definite integral differs from an indefinite integral in that it produces a numerical value, encompassing the accumulation of quantities represented by the function over the interval. In the calculation of the average value of a function, the definite integral \( \int_0^3 \frac{x}{\sqrt{x^2+16}} \, dx \) is used to determine the total area under the curve of \( f(x) = \frac{x}{\sqrt{x^2+16}} \) from \( x=0 \) to \( x=3 \). This is then divided by the interval's length to obtain an average.

Substitution Method

The substitution method is a useful technique in calculus for solving complex integrals by simplifying them into an easier form. The main idea is to replace a part of the integrand with a single variable, simplifying the integral.

• First, choose a substitution \( u \) such as \( u = x^2 + 16 \), simplifying the expression inside the integrand.
• Then, find \( du = 2x \, dx \) which allows replacing \( x \, dx \) with \( \frac{du}{2} \).
• Subsequently, rewrite the original integral in terms of \( u \) making it easier to solve, leading to \( \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \).

The substitution method is particularly helpful when dealing with integrals involving composite functions, where direct integration is cumbersome.

Limits of Integration

The limits of integration are vital as they define the start and end points of the definite integral. In a problem, such as the one given, it starts at \( x = 0 \) and ends at \( x = 3 \).

• Upon substitution, these limits must be adjusted to match the new variable. For \( u = x^2 + 16 \), when \( x = 0 \), \( u = 16 \); and when \( x = 3 \), \( u = 25 \).
• These new limits ensure the calculation is performed over the correct interval relative to the substitution variable \( u \).

Choosing and adjusting limits accurately ensures the definite integral evaluates the correct area over the desired range.

Continuous Function

A continuous function is a type of mathematical function where small changes in input result in small changes in output, without any breaks or gaps. This property is crucial for calculating definite integrals, as the integration process involves summing up infinitely small values across an interval.For the specific function \( f(x) = \frac{x}{\sqrt{x^2+16}} \), the continuity over the interval \([0, 3]\) allows for the proper application of integration techniques such as substitution. Without continuity, the integrals could yield incorrect representations of area, emphasizing why ensuring continuity is a step usually verified in such problems.