Problem 111 - Calculus Volume 1

Question

In the following exercises, find the average value \(f_{\text { ave }}\) of \(f\) between \(a\) and \(b,\) and find a point \(c,\) where \(f(c)=f_{\text { ave. }}\) \(f(x)=x^{5}, a=-1, b=1\)

Step-by-Step Solution

Answer

The average value is 0; the point where the function equals this average is at \( c = 0 \).

Step 1: Determine the Average Value Formula

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by \( f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). For this problem, \( a = -1 \) and \( b = 1 \).

Step 2: Set Up the Integral

Write the integral that needs to be evaluated: \( \int_{-1}^{1} x^5 \, dx \).

Step 3: Compute the Integral

To find the integral, calculate \( \int x^5 \, dx = \frac{x^6}{6} + C \). Evaluate this expression from \( -1 \) to \( 1 \).

Step 4: Evaluate the Definite Integral

Evaluate \( \frac{x^6}{6} \) at \( x = 1 \) and \( x = -1 \), and subtract: \[ \left( \frac{1^6}{6} \right) - \left( \frac{(-1)^6}{6} \right) = \frac{1}{6} - \frac{1}{6} = 0 \].

Step 5: Calculate the Average Value

To find \( f_{\text{ave}} \), divide the integral result by \( b-a \): \( f_{\text{ave}} = \frac{0}{1 - (-1)} = \frac{0}{2} = 0 \).

Step 6: Find Point \( c \) such that \( f(c) = f_{\text{ave}} \)

Set \( f(c) = c^5 = 0 \). Solve for \( c \), which means \( c = 0 \).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus that represents the signed area under a curve. It helps us find the accumulation of quantities, like area, volume, and even total distance. When we talk about the definite integral of a function \( f(x) \) over an interval \([a, b]\), it is mathematically expressed as:

• \( \int_{a}^{b} f(x) \, dx \).

This notation signifies the summation of infinitely small sections of the area between the curve and the x-axis, from \( x = a \) to \( x = b \). In practical terms, it finds out how much total "stuff" is accumulated across this segment of the x-axis.
To compute it, one takes the antiderivative of the function and evaluates it at the boundary limits \( a \) and \( b \). Then, by subtracting these results, you obtain the definite integral's value. This step is key when discovering the average value of a function on a specific interval.

Evaluating Integrals

Evaluating integrals refers to the process of finding the integral's value. Especially with definite integrals, you perform this by calculating the antiderivative and plugging in the interval's limits.

• For example, the integral \( \int x^5 \, dx \) becomes \( \frac{x^6}{6} + C \), where \( C \) is a constant which cancels out in definite integrals.

In our task, we evaluate this expression between \( x = -1 \) and \( x = 1 \), which gives:

• \[ \left( \frac{1^6}{6} \right) - \left( \frac{(-1)^6}{6} \right) = \frac{1}{6} - \frac{1}{6} = 0 \]

Evaluating this integral verified the average value integral's role, since a symmetric interval and an odd power of \( x \) (in this case, the fifth power) result in zero area. Thus, confirming that the integral's value influences the determination of a function's average value.

Finding Points on a Function

After determining the average value of the function over an interval, the next step is to find a point \( c \) on the function where the function's output equals this average value. If the average value of \( f(x) \), represented as \( f_{\text{ave}} \), was calculated to be zero, then we solve:

• \( f(c) = c^5 = 0 \).

This equation simply points to the value of \( c \) where \( c^5 = 0 \). Solving for \( c \) reveals that \( c = 0 \). Thus, point \( c = 0 \) in our example is the location on the curve \( f(x) = x^5 \) where the function’s value coincides with its computed average value over the interval \([-1, 1]\).
Identifying such points helps in understanding the general behavior of a function across a given segment and provides deeper insights into its geometric representation.