Question

Find the average value of the function on the given interval. $$ h(z)=\frac{\sin \sqrt{z}}{\sqrt{z}} ; \quad[\pi / 4, \pi / 2] $$

Short answer

The average value of the function on the interval is \( \frac{8}{\pi}(-\cos(\sqrt{\pi/2}) + \cos(\sqrt{\pi}/2)) \).

Step-by-step solution

Write the formula for the average value of a function

The average value \( f_{avg} \) of a function \( f(x) \) on an interval \([a, b]\) is given by the formula:\[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx\]Apply this formula to the function \( h(z) = \frac{\sin \sqrt{z}}{\sqrt{z}} \) on the interval \( [\pi/4, \pi/2] \).

Set up and simplify the integral

Substitute \( h(z) = \frac{\sin \sqrt{z}}{\sqrt{z}} \) and the limits of integration into the formula:\[ f_{avg} = \frac{1}{\pi/2 - \pi/4} \int_{\pi/4}^{\pi/2} \frac{\sin \sqrt{z}}{\sqrt{z}} \, dz\]Simplify the denominator:\[ \pi/2 - \pi/4 = \pi/4\]So the formula becomes:\[ f_{avg} = \frac{4}{\pi} \int_{\pi/4}^{\pi/2} \frac{\sin \sqrt{z}}{\sqrt{z}} \, dz\]

Make a substitution to evaluate the integral

Let \( u = \sqrt{z} \), so \( z = u^2 \) and \( dz = 2u \, du \). Change the limits of integration: \( z = \pi/4 \Rightarrow u = \sqrt{\pi/4} = \sqrt{\pi}/2 \) and \( z = \pi/2 \Rightarrow u = \sqrt{\pi/2} \).Substitute into the integral:\[ \int_{\sqrt{\pi}/2}^{\sqrt{\pi/2}} \sin u \, du = \int_{\sqrt{\pi}/2}^{\sqrt{\pi/2}} \sin u \, 2u \, du\]Since \( \sqrt{z} = u \), it simplifies to:\[ 2 \int_{\sqrt{\pi}/2}^{\sqrt{\pi/2}} \sin u \, du\]

Solve the integral

The integral \( \int \sin u \, du \) is \(-\cos u\). Thus:\[ 2 \int_{\sqrt{\pi}/2}^{\sqrt{\pi/2}} \sin u \, du = 2 \left[ -\cos u \right]_{\sqrt{\pi}/2}^{\sqrt{\pi/2}}\]Evaluate:\[ 2 \left( -\cos(\sqrt{\pi/2}) + \cos(\sqrt{\pi}/2) \right)\]

Evaluate and simplify the result

Calculate the cosine values and simplify:\[ 2 \left( -\cos(\sqrt{\pi/2}) + \cos(\sqrt{\pi}/2) \right)\]Since these are exact values, retain the expression as a representation of the average value using numerical methods or approximation if needed.

Calculate the average value

Multiply by the fraction from the average value formula:\[ f_{avg} = \frac{4}{\pi} \cdot 2 \left( -\cos(\sqrt{\pi/2}) + \cos(\sqrt{\pi}/2) \right)\]This is the expression for the average value of \( h(z) \) over the interval \([\pi/4, \pi/2]\).

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus used to find areas, volumes, and other quantities. Various techniques can make solving integrals easier, especially complex ones. One such technique is the use of definite integrals, which provide a precise value for the integral over a specific interval. In this exercise, we use one of the integration techniques to find the average value of the function. The core integration technique here involves setting up the integral using the standard formula for the average value of a continuous function. This approach lets us determine the average or mean value of the given function over its specified range. By applying definite integrals, we calculate the overall behavior of the function within a specific domain, giving us insight into its average trend.

Trigonometric Functions

Trigonometric functions like sine and cosine arise frequently in calculus problems and are essential in many fields of mathematics. In this exercise, the function involves the sine of a square root, specifically \[h(z) = \frac{\sin \sqrt{z}}{\sqrt{z}}.\]Understanding trigonometric functions and how to integrate them is critical for solving calculus problems. The sine function poses unique challenges due to its oscillating nature, but it also has well-known integration and differentiation properties. Here, we leverage these properties to compute the integral by transforming the original function into a simpler form for integration.

Substitution Method

The substitution method, often referred to as "u-substitution," is a powerful tool for solving integrals that are not straightforward. In essence, it simplifies an integral by substituting a part of the original function with a new variable, making the integral easier to evaluate.In this problem, we use the substitution \[u = \sqrt{z}\],which transforms the integral into a form that uses the variable \(u\). Through this method, we replace complex expressions, allowing us to work with simpler integrals involving basic trigonometric functions. The substitution often involves adjusting the differential and limits of integration as we do here, making it essential to understand both the mechanics and logic behind substitutions thoroughly.

Definite Integrals

Definite integrals provide the net area under the curve of a function between two points, giving us more than just a solution—they offer insight into the behavior of functions over specific intervals. When solving for the average value of a function over an interval, definite integrals come into play by allowing us to calculate a finite area, rather than an indefinite one.In this exercise, the definite integral has boundaries defined by the interval \[\left[\frac{\pi}{4}, \frac{\pi}{2}\right].\]The average value is derived by evaluating the definite integral of the function over this range. This process involves careful calculation of both the integral and the interval size to determine the mean behavior of the function over the specified domain, essentially allowing us to "average out" its effects.