Problem 53 - Calculus

Question

Assuming that \(u\) and \(v\) can be integrated over the interval \([a, b]\) and that the average values over the interval are denoted by \(\bar{u}\) and \(\bar{v},\) prove or disprove that (a) \(\bar{u}+\bar{v}=\overline{u+v}\) (b) \(k \bar{u}=\overline{k u},\) where \(k\) is any constant; (c) if \(u \leq v\) then \(\bar{u} \leq \bar{v}\).

Step-by-Step Solution

Answer

All statements (a), (b), and (c) are true.

Step 1: Understand Average Value

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by the formula \( \bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx \). We need to apply this concept for the functions \(u\) and \(v\) to evaluate the statements.

Step 2: Evaluate Statement (a) \( \bar{u} + \bar{v} = \overline{u + v} \)

Calculate the left side: \(\bar{u} + \bar{v} = \frac{1}{b-a} \int_a^b u(x) \, dx + \frac{1}{b-a} \int_a^b v(x) \, dx = \frac{1}{b-a} \left( \int_a^b u(x) \, dx + \int_a^b v(x) \, dx \right).\)Calculate the right side: \(\overline{u+v} = \frac{1}{b-a} \int_a^b (u(x) + v(x)) \, dx = \frac{1}{b-a} \left( \int_a^b u(x) \, dx + \int_a^b v(x) \, dx \right).\)Since both sides are equal, the statement is proved true.

Step 3: Evaluate Statement (b) \( k \bar{u} = \overline{k u} \)

Calculate the left side: \(k \bar{u} = k \cdot \frac{1}{b-a} \int_a^b u(x) \, dx.\)Calculate the right side: \(\overline{k u} = \frac{1}{b-a} \int_a^b k u(x) \, dx = k \cdot \frac{1}{b-a} \int_a^b u(x) \, dx.\)Since both sides are the same, this statement is also true.

Step 4: Evaluate Statement (c) \( u \leq v \Rightarrow \bar{u} \leq \bar{v} \)

If \(u(x) \leq v(x)\) for all \(x\) in \([a, b]\), then \(\int_a^b u(x) \, dx \leq \int_a^b v(x) \, dx.\)Thus, \( \frac{1}{b-a} \int_a^b u(x) \, dx \leq \frac{1}{b-a} \int_a^b v(x) \, dx,\) which implies \(\bar{u} \leq \bar{v}.\)Therefore, this statement is also true.

Key Concepts

Average Value of a Function

When you think about the average value of a function, imagine taking a smooth hill and flattening it out into a rectangle. The average value is like the height of that rectangle. Mathematically, this is expressed as the mean height over a given interval \([a, b]\). The formula we use is: \[\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx\]This equation tells you how to handle functions like \(u\) and \(v\) over an interval. It's beneficial when analyzing each scenario separately and collectively, as shown in our original exercise.

• For two functions like \(u\) and \(v\): their combined average is simply the sum of their individual averages, \(\bar{u} + \bar{v} = \overline{u+v}\).
• Scaling a function by a constant \(k\) also scales its average value: \(k \bar{u} = \overline{ku}\).

In simpler terms, integrating on an interval shows how much of the function value is spread over the interval, and the average just distributes it evenly across that interval.

Inequality in Integrals

Integrals capture how a function behaves over an interval, almost like a snapshot of its 'volume'. When comparing two functions, for instance, if you know \(u(x) \leq v(x)\) across every point in \([a, b]\), then:

• The integral of \(u\) will be less than or equal to the integral of \(v\).
• This makes sense because everywhere you have \(u\) under \(v\), the area \(u\) covers will be smaller or similar.

This property allows us to state that if a function \(u(x)\) never exceeds \(v(x)\), its average value will always be lesser or equal. Thus, \(\bar{u} \leq \bar{v}\). Understanding this concept is crucial as it relates to inequalities found in integral solutions.

Properties of Integrals

Integrals come with some flexible rules that make calculations both straightforward and powerful. Here, we'll highlight why these properties are practical:

• Additivity: If you integrate two functions separately and add their results, it's equivalent to integrating their sum. For instance, \(\int_a^b (u(x) + v(x)) \, dx = \int_a^b u(x) \, dx + \int_a^b v(x) \, dx\).
• Linearity: Multiplying a function by a constant also multiplies its integral by that constant. So, \(\int_a^b k \cdot u(x) \, dx = k \cdot \int_a^b u(x) \, dx\).

These properties simplify the computation of average values and verifying inequalities. They are handy for verifying statements about functions and their average behavior, as we've seen in our original exercise, supporting proofs with robust mathematical thinking.