Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the average value of the function on the given interval. $$ f(x)=\sin x ;[0, \pi] $$

Step-by-Step Solution

Answer

The average value is \( \frac{2}{\pi} \).

Step 1: Understand the Formula for Average Value of a Function

The average value of a function \( f(x) \) over an interval \([a, b]\) is given by the formula:\[avg = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx\]For our function \( f(x) = \sin x \) on the interval \([0, \pi]\), this formula becomes:\[avg = \frac{1}{\pi - 0} \int_{0}^{\pi} \sin x \, dx\]

Step 2: Set Up the Integral

We need to set up the integral based on the function \( f(x) = \sin x \) and interval \([0, \pi]\):\[\int_{0}^{\pi} \sin x \, dx\]This represents the definite integral of \( \sin x \) from 0 to \( \pi \).

Step 3: Solve the Integral

To solve \( \int_{0}^{\pi} \sin x \, dx \), recall the integral of sine:\[\int \sin x \, dx = -\cos x + C\]Plug in the bounds 0 and \( \pi \):\[[-\cos x]_{0}^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2\]

Step 4: Calculate the Average Value

Now, use the result of the integral to find the average value:\[avg = \frac{1}{\pi} \times 2 = \frac{2}{\pi}\]

Step 5: Interpret the Result

The average value of the function \( f(x) = \sin x \) over the interval \([0, \pi]\) is \( \frac{2}{\pi} \). This value represents the mean height of the sine wave over this interval.

Key Concepts

Definite Integral

A definite integral represents the accumulation of quantities, such as areas under curves. It provides a numerical value instead of a function like indefinite integrals do. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). This calculates the area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
Key properties of definite integrals include:

• Additivity: \( \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx \)
• Linearity: Scaling by a constant can be factored out, i.e., \( \int_{a}^{b} cf(x) \, dx = c \int_{a}^{b} f(x) \, dx \).

To calculate a definite integral, you can use the Fundamental Theorem of Calculus. This states that if \( F(x) \) is an antiderivative of \( f(x) \), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]In the solution provided, calculating \( \int_{0}^{\pi} \sin x \, dx \) involves recognizing \( -\cos x + C \) as the indefinite integral of \( \sin x \) and then applying the limits \( x = 0 \) and \( x = \pi \).

Sine Function

The sine function is a fundamental trigonometric function represented as \( \sin x \). It periodic and oscillates between -1 and 1. The sine function is critical in describing waveforms and circular motion.
Important characteristics of the sine function include:

• Periodicity: \( \sin x \) has a period of \( 2\pi \), meaning \( \sin(x + 2\pi) = \sin x \).
• Symmetry: It is an odd function, which implies \( \sin(-x) = -\sin x \).
• Zeros: It crosses the x-axis at integer multiples of \( \pi \) (i.e., \( x = n\pi \), where \( n \) is an integer).

In the context of the exercise, the sine function within the given interval \([0, \pi]\) first increases from 0 to 1 and then decreases back to 0. Calculating the average value over this interval essentially finds a mean value that represents the overall height of the curve over the interval.

Integration by Substitution

Integration by substitution is a technique for evaluating integrals and is similar to the chain rule for differentiation. The method involves substituting a part of the integrand with a single variable, simplifying the integration process.
Let's understand some basic steps of integration by substitution:

• Step 1: Choose a substitution \( u = g(x) \), derive \( du = g'(x)dx \).
• Step 2: Rewrite the integral using \( u \) and \( du \). This changes the original integral into a simpler form.
• Step 3: Integrate with respect to \( u \), then substitute back the original variable.

This technique isn't directly used in the provided solution since \( \int \sin x \, dx \) is already standard. However, understanding substitution helps with more complex integrals by turning an intimidating integral into a more straightforward calculation.