Problem 69 - Calculus Single Variable

Question

In Exercises \(69-72,\) compute the average value \(f_{\text {avg }}\) of \(f\) over \([a, b]\), and find a value of \(c\) in \((a, b)\) at which \(f\) attains this average value. Illustrate the geometric meaning of the Mean Value Theorem for Integrals with a graph. $$ f(x)=x+\sin (x) \quad a=0, b=\pi / 2 $$

Step-by-Step Solution

Answer

The average value is \(f_{\text{avg}} = \frac{\pi}{4} + \frac{2}{\pi}\). Solve \(c + \sin(c) = \frac{\pi}{4} + \frac{2}{\pi}\) for \(c\).

Step 1: Understand the Formula for Average Value

The average value of a function \(f\) on the interval \([a, b]\) is given by \(f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). We will use this formula to find the average value of the function \(f(x) = x + \sin(x)\) over \([0, \pi/2]\).

Step 2: Set Up the Definite Integral

First, substitute \(f(x) = x + \sin(x)\), \(a = 0\), and \(b = \pi/2\) into the formula for the average value. This gives\[ f_{\text{avg}} = \frac{1}{\pi/2 - 0} \int_{0}^{\pi/2} (x + \sin(x)) \, dx \].

Step 3: Compute the Integral

Compute \( \int_{0}^{\pi/2} (x + \sin(x)) \, dx = \int_{0}^{\pi/2} x \, dx + \int_{0}^{\pi/2} \sin(x) \, dx \).1. \(\int x \, dx\), the antiderivative of \(x\) is \(\frac{x^2}{2}\). So,\[ \left[\frac{x^2}{2}\right]_{0}^{\pi/2} = \frac{(\pi/2)^2}{2} - \frac{0^2}{2} = \frac{\pi^2}{8}.\]2. \(\int \sin(x) \, dx\), the antiderivative of \(\sin(x)\) is \(-\cos(x)\). So,\[ \left[-\cos(x)\right]_{0}^{\pi/2} = (0 - (-1)) = 1.\]

Step 4: Calculate the Average Value

Add the values from the integrals:\[ \int_{0}^{\pi/2} (x + \sin(x)) \, dx = \frac{\pi^2}{8} + 1.\]Substitute this result back to find the average value:\[ f_{\text{avg}} = \frac{1}{\pi/2} \left(\frac{\pi^2}{8} + 1\right) = \frac{2}{\pi} \left(\frac{\pi^2}{8} + 1\right).\]Simplifying gives:\[ f_{\text{avg}} = \frac{\pi}{4} + \frac{2}{\pi}.\]

Step 5: Find the Value of \(c\)

To find \(c\) such that \(f(c) = f_{\text{avg}}\), solve \(c + \sin(c) = \frac{\pi}{4} + \frac{2}{\pi}\). This usually requires numerical methods or graphing tools, such as using a calculator to approximate the solution.

Step 6: Illustrate the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that there exists a value \(c\) in \((a, b)\) such that \(f(c) = f_{\text{avg}}\). On a graph of \(f(x)\) over \([0, \pi/2]\), draw a horizontal line at \(f_{\text{avg}}\). The point where this line intersects the graph represents the value \(c\).

Key Concepts

Average Value of a Function

The average value of a function over a specific interval gives us insight into the function's behavior across that stretch. To calculate this, we use a formula that involves integration. The formula to find the average value of a function \( f(x) \) over the interval \([a, b]\) is:

• \( f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

This formula takes the integral of the function over the specified interval and divides it by the length of the interval. By calculating this average, we essentially find the constant value that the function averages out to over \([a, b]\). For the function \( f(x) = x + \sin(x) \) on \([0, \pi/2]\), the average value was found as \( \frac{\pi}{4} + \frac{2}{\pi} \), representing a balance between the function's peaks and troughs.

Definite Integral

The definite integral is a vital concept for understanding how to find the average value of a function. It measures the total accumulation of a function over an interval \([a, b]\). Using integration, you can sum up a continuous range of values that a function yields.

• For \( \int_{0}^{\pi/2} (x + \sin(x)) \, dx \), this means calculating the area under the curve from \( x = 0 \) to \( x = \pi/2 \).
• This involves integrating the function in parts; both \(x\) and \(\sin(x)\) are integrated separately.

After calculating, the area, or the net signed area for \( f(x) = x + \sin(x) \), resulted in \(\frac{\pi^2}{8} + 1\). This computation sums up the values closely tied to the essence of the Mean Value Theorem for Integrals.

Function Graphing

Graphing a function is a visual tool that helps in understanding the relationship between variables. It turns mathematical expressions into visual representations. For the Mean Value Theorem for Integrals, graphing \( f(x) \) allows us to see where the average value line intersects the curve.

• Imagine the graph of \( f(x) = x + \sin(x) \) over \([0, \pi/2]\).
• By drawing a horizontal line at \( f_{\text{avg}} = \frac{\pi}{4} + \frac{2}{\pi} \), we visualize exactly where this line intersects with \( f(x) \).

This intersection point is crucial; it tells us the \(c\) value satisfying \(f(c) = f_{\text{avg}}\). Seeing the graph aids in linking abstract calculations with tangible outcomes.

Numerical Methods

To locate precise values of \(c\) where \(f(c) = f_{\text{avg}}\), numerical methods become indispensable. These techniques allow us to approximate solutions for equations that do not resolve easily through algebra alone.

• Methods like the Bisection Method, Newton's Method, or simply graphic calculator approximations help solve \(c + \sin(c) = \frac{\pi}{4} + \frac{2}{\pi}\).
• These methods iteratively refine guesses until a practical \(c\) value emerges.

Because exact algebraic solutions might be impractical or impossible, numerical methods provide the practical means to solve such equations. They bridge the gap between theoretical mathematics and applied problem-solving.