Problem 54 - Calculus: Early Transcendental Functions

Question

Suppose that the average value of a function \(f(x)\) over an interval \([a, b]\) is \(v\) and the average value of \(f(x)\) over the interval [b, \(c]\) is \(w .\) Find the average value of \(f(x)\) over the interval \([a, c]\).

Step-by-Step Solution

Answer

The average value of \(f(x)\) over the interval [a, c] is \((v(a - c) + w(c - b)) / (c - a)\).

Step 1: Calculation of the average value of a function over the intervals [a, b] and [b, c]

The average value \(v\) of \(f(x)\) over the interval [a, b] is obtained by evaluating \((1/(b-a)) \int_{a}^{b} f(x) dx \) , and the average value \(w\) of \(f(x)\) over the interval [b, c] is obtained by evaluating \((1/(c-b)) \int_{b}^{c} f(x) dx\). This is given in the problem.

Step 2: Applying the formula for the combined interval [a, c]

To find the average value over the combined interval [a, c], we need to apply the formula for the average value of a function on the interval [a, c], which is \((1/(c-a)) \int_{a}^{c} f(x) dx\).

Step 3: Equating both equations and solving for the average value over [a, c]

Using the above expressions, \((1/(c-a)) \int_{a}^{c} f(x) dx = v(b - a) / (c -a) + w(c - b) / (c - a)\), we can simplify to get \(v(a - c) /(c - a) + w\).

Key Concepts

Definite Integral

To understand the average value of a function over an interval, we need to first comprehend the concept of the definite integral. The definite integral of a function from point \(a\) to point \(b\) is essentially the accumulation of the function's values over the interval \([a, b]\). It is represented mathematically as \(\int_{a}^{b} f(x) \, dx\). This integral calculates the total area under the curve \(f(x)\) from \(a\) to \(b\).
\[ \int_{a}^{b} f(x) \, dx \text{ gives us the total "accumulation", or "area" under } f(x) .\]

• Used to determine the overall "effect" of \(f(x)\) over a specific interval.
• Provides a means to calculate the average value by dividing the integral by the length of the interval.

The formula for the average value of a function using the definite integral is given by:\[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]
This equation divides the total area by the interval length \((b-a)\), translating to an average "height" of the function in that region.

Intervals in Calculus

In calculus, intervals are portions of the real number line between two end points. They are fundamental in analyzing functions, especially when calculating definite integrals.
When we mention an interval such as \([a, b]\), it includes all numbers between \(a\) and \(b\). There are different types of intervals:

• Closed interval \([a, b]\) includes both endpoints \(a\) and \(b\).
• Open interval \((a, b)\) does not include \(a\) or \(b\).
• Half-open interval \([a, b)\) includes \(a\) but not \(b\), or \((a, b]\) which includes \(b\) but not \(a\).

Understanding these intervals is crucial when calculating integrals and average values over specific sections of a function. It defines precisely the area where the function is evaluated.

Function Analysis

Analyzing a function involves understanding its behavior over a given interval. When given functions such as \(f(x)\), it's essential to analyze their characteristics like continuity, maximum, minimum, and average value over specified intervals.
For instance, if you have a function and you're given its average values over different sections like \([a, b]\) and \([b, c]\), these might have differing impacts based on the function's behavior over those intervals.
To find the overall average value of the function over a larger interval \([a, c]\), you can combine these individual analyses. Using the weighted formula:\[ \text{Average over } [a, c] = \frac{1}{c-a} \left(v(b-a) + w(c-b)\right)\] This combination accounts for the size of each interval contributing to the total.
In essence:

• An average value represents how the function "behaves" as an overall "height" over the interval.
• Exploring the function's actual values and patterns helps interpret the accumulated changes over an interval.
• Understanding average value calculations is key in predicting the general behavior of the function over extended ranges.

With this knowledge, interpreting function analysis allows for a deeper grasp of any given problem involving function behavior over specific ranges.