Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(0 \leq f(x) \leq g(x)\) for all \(x\) in \([0,1]\), and let \(R\) and \(S\) be the regions under the graphs of \(f\) and \(g\), respectively. Prove or disprove that \(\bar{y}_{R} \leq \bar{y}_{S}\).

Step-by-Step Solution

Answer

Yes, \(\bar{y}_R \leq \bar{y}_S\) because \(0 \leq f(x) \leq g(x)\) implies \(\int_0^1 f(x) \, dx \leq \int_0^1 g(x) \, dx\).

Step 1: Understanding the problem

We are given two functions, \(f(x)\) and \(g(x)\), such that \(0 \leq f(x) \leq g(x)\) over the interval \([0,1]\). We need to prove or disprove that the average value (height) of \(f(x)\) is less than or equal to the average value of \(g(x)\) over this interval.

Step 2: Finding the average values

The average value of a function \(h(x)\) over an interval \([a,b]\) is given by \(\bar{y}_h = \frac{1}{b-a} \int_a^b h(x) \, dx\). So, \(\bar{y}_{R} = \int_0^1 f(x) \, dx\) and \(\bar{y}_{S} = \int_0^1 g(x) \, dx\), since the length of the interval is 1.

Step 3: Using the property of the integrals

Since \(f(x)\) is always less than or equal to \(g(x)\) over \([0,1]\), we can use the property of definite integrals: If \(a(x) \leq b(x)\) for all \(x\) in the interval, then \(\int_a^b a(x) \, dx \leq \int_a^b b(x) \, dx\). Applying this property, we get \(\int_0^1 f(x) \, dx \leq \int_0^1 g(x) \, dx\).

Step 4: Concluding the proof

Since \(\int_0^1 f(x) \, dx \leq \int_0^1 g(x) \, dx\), it directly follows that \(\bar{y}_R \leq \bar{y}_S\). Therefore, the statement is proven to be true.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus used to calculate the accumulation of quantities, such as area under a curve. When calculating a definite integral, we look at the area of a region bounded by a function. This operation transforms a function into a single number that represents this area.
In the context of this exercise, definite integrals help us find the 'average value' or average height of the functions over the given interval. For instance, if we want to determine the average value of a function \(h(x)\) over an interval \([a, b]\), we use the formula:

• \( \bar{y}_h = \frac{1}{b-a} \int_a^b h(x) \, dx \)

The calculation involves integrating the function over the interval and then dividing by the length of that interval. This gives us a single number that represents the mean function value over \([a, b]\). Through definite integrals, we can evaluate, for example, how much more one function can rise above another over a specific interval.

Inequalities

Inequalities are an essential tool in mathematics. They show relationships when one expression is bigger or smaller than another. For example, saying \( f(x) \leq g(x) \) means that for every value of \(x\) in our interval, \(f(x)\) is not larger than \(g(x)\).
This principle is very useful when dealing with integrals. If \(a(x) \leq b(x)\) for all \(x\) in \([a, b]\), then the integral of \(a(x)\) over this interval will also be less than or equal to the integral of \(b(x)\).
This property, applied to our functions \(f(x)\) and \(g(x)\), directly helps us in proving that one average value cannot exceed the other. By ensuring the inequality holds throughout their interval, we can safely say that the integral of one function won't overtake that of a larger one, making comparisons and evaluations between the two dependable.

Functions Over Intervals

Functions over specific intervals are crucial in understanding many aspects of calculus and algebra. An interval like \([0, 1]\) defines a specific range for the variable \(x\) within which we evaluate functions.
Focusing on functions within intervals lets us closely examine behavior and characteristics that might not be visible otherwise. When examining \(f(x)\) and \(g(x)\) between 0 and 1, we consider all possible values for \(x\) within this range.
Using such intervals, we can apply tests like checking inequalities or finding average values. We're able to compare exact regions under curves using definite integrals, which in turn gives us practical insights into how functions relate to one another. Therefore, dissecting functions over intervals is fundamental when evaluating their respective properties, such as how much area they encompass, potential maxima, and minima, or average rates over set periods.