Question

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

Short answer

The average value of the function is 0.

Step-by-step solution

Understand the Problem

To find the average value of a continuous function over an interval, we use the formula \[ f_{ ext{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \]. For our function \( f(x) = \sin^2 x \cos^3 x \), we will evaluate this integral over the interval \([-\pi, \pi]\).

Set Up the Integral

Using the formula for the average value of a function, set up the integral: \[ f_{ ext{avg}} = \frac{1}{2i ext{-}(-2)} \int_{-2}^{2} \sin^2 x \cos^3 x \, dx .\] This simplifies to \( f_{ ext{avg}}n=n\frac{1\1} \int_{-2}^{2} \sin^2 x \cos^3 x \, dx \).

Simplify Using Symmetry

Notice that \( f(x) = \sin^2 x \cos^3 x \) is an odd function because \( \sin^2(-x)\cos^3(-x) = -\sin^2(x)\cos^3(x) \). Therefore, the integral over a symmetric interval around zero \([-2, 2])\) of an odd function is zero. Thus, \( \int_{-2}^{2} \sin^2 x \cos^3 x \, dx = 0 \).

Calculate the Average Value

Since the integral evaluates to zero due to the function's symmetry, the average value is \( f_{ ext{avg}} = \frac{1}{2\pi} \times 0 = 0 \).

Key concepts

Definite Integral

In mathematics, a definite integral plays a crucial role in finding the total accumulation of quantities, such as area under a curve over a specific interval. Given a continuous function \( f(x) \), the definite integral \( \int_a^b f(x) \, dx \) calculates the net area between the graph of the function and the x-axis from \(x = a\) to \(x = b\). This calculation requires knowing the endpoints of integration, \(a\) and \(b\), and the function itself.

The main properties of definite integrals include:

• Linearity: The integral of a sum is equal to the sum of the integrals, and constants can be factored out.
• Interval Reversal: Reversing the limits of integration changes the sign of the integral.
• Zero-width Interval: The integral over an interval with no width is zero.

To find the average value of a function, the integral is used in the formula:
\[ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \]
This allows us to distribute the "total accumulation" evenly over the interval from \(a\) to \(b\). Understanding definite integrals enables us to solve problems involving accumulated change and average values.

Continuous Function

A continuous function is one that does not have any breaks, holes, or jumps in its graph within a given interval. Essentially, you can draw its graph without lifting your pen from the paper. Mathematically, a function \( f(x) \) is continuous on an interval if for every point \( c \) in the interval, \( \lim_{x \to c} f(x) = f(c) \).

Continuous functions have several important properties, including:

• The Intermediate Value Theorem: If a function is continuous on a closed interval \([a, b]\), it takes every value between \( f(a) \) and \( f(b) \).
• Integrable: Continuous functions are always integrable over a closed interval, meaning we can compute the definite integral.
• Predictable Behavior: Because of their nature, continuous functions generally behave in a smooth and predictable manner.

In the context of finding an average value over an interval, continuity ensures that the function behaves well across the interval, allowing for accurate computation of the integral and thereby the average value.

Odd Function

An odd function is a type of function characterized by its symmetry about the origin. That is, a function \( f(x) \) is considered odd if for all \( x \) in its domain, \( f(-x) = -f(x) \). This peculiar property greatly affects the behavior of the function's integral over symmetric intervals.

Here are the key points about odd functions:

• Symmetry about Origin: Graphically, each part of the function has a corresponding part in the opposite quadrant.
• Integral over Symmetric Interval: When you integrate an odd function over a symmetric interval around zero, like \([-a, a]\), the result is zero.

This zero-result property is especially useful in simplifying complex integrals. As shown in the solution, knowing that \( \sin^2 x \cos^3 x \) is odd helped in concluding the integral from \(-\pi\) to \(\pi\) as zero, thus making the calculation of the average value straightforward.

Symmetry in Calculus

Symmetry in calculus refers to the balanced and mirrored properties of functions that can greatly simplify the evaluation of integrals and understanding of function behavior. When a function exhibits symmetry, it shows similar properties either about the y-axis, x-axis, or origin, which helps in reducing complex calculations.

Different types of symmetry include:

• Even Symmetry: Functions have even symmetry if \( f(-x) = f(x) \); their graphs are symmetric about the y-axis.
• Odd Symmetry: Functions have odd symmetry if \( f(-x) = -f(x) \); their graphs are symmetric about the origin.

Recognizing symmetry provides valuable insights:

• Integral Simplification: As demonstrated, for odd functions over symmetric intervals, the integral is zero, simplifying calculations.
• Predictive Analysis: Symmetry can aid in predicting function outputs and graph shapes, which is crucial for analysis.

In the discussed problem, the symmetry of the function allowed us to notice quickly that the integral of \( \sin^2 x \cos^3 x \) over \([-\pi, \pi]\) was zero, relying on the symmetry properties of odd functions in a symmetrical interval.