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 of disprove that \(\bar{y}_{R} \leq \bar{y}_{S}\).

Short answer

The statement \(\bar{y}_R \leq \bar{y}_S\) is true.

Step-by-step solution

Understand the Problem

We are given two functions, \(f(x)\) and \(g(x)\), with the condition \(0 \leq f(x) \leq g(x)\) for all \(x\) in \([0,1]\). We need to compare the average values of these functions over this interval, represented as \(\bar{y}_{R}\) for \(f(x)\) and \(\bar{y}_{S}\) for \(g(x)\).

Express Average Values in Integral Form

The average value of a function \(h(x)\) over an interval \([a,b]\) is given by \(\bar{y} = \frac{1}{b-a} \int_{a}^{b} h(x) \,dx\). Therefore, \(\bar{y}_{R} = \int_{0}^{1} f(x) \,dx\) and \(\bar{y}_{S} = \int_{0}^{1} g(x) \,dx\), since both are over the interval \([0,1]\).

Use Given Function Inequality to Compare Integrals

Given \(0 \leq f(x) \leq g(x)\), by the properties of definite integrals, it follows that \(\int_{0}^{1} f(x) \,dx \leq \int_{0}^{1} g(x) \,dx\). This implies that the area under \(f\) is less than or equal to the area under \(g\).

Analyze the Average Values

Since \(\bar{y}_R = \int_{0}^{1} f(x) \,dx\) and \(\bar{y}_S = \int_{0}^{1} g(x) \,dx\), and from Step 3 we have \(\int_{0}^{1} f(x) \,dx \leq \int_{0}^{1} g(x) \,dx\), it follows that \(\bar{y}_R \leq \bar{y}_S\).

Conclusion

Given the inequality \(\int_{0}^{1} f(x) \,dx \leq \int_{0}^{1} g(x) \,dx\), we conclude that the average value of \(f(x)\) over \([0,1]\) is less than or equal to the average value of \(g(x)\) over the same interval. Thus, \(\bar{y}_R \leq \bar{y}_S\) is proven to be true.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in integral calculus, used to find the accumulated value of a function over a specific interval. Imagine you have a curve represented by a function, and you want to find the total area beneath it between two points on the x-axis, say from point \(a\) to point \(b\). That's where definite integrals come into play. You write this as \(\int_{a}^{b} f(x) \, dx\).
The definite integral gives not only the net area but actually the accumulated sum of all these tiny little slivers underneath the curve.
This makes it extremely useful in solving problems where accumulation is involved, like finding distances, areas, and even probabilities.
In our exercise, we use definite integrals to compare the total areas under two different functions over the same interval. Since one function \(f(x)\) is less than or equal to the other function \(g(x)\), the integral of \(f(x)\) over \([0, 1]\) will be less than or equal to the integral of \(g(x)\) over the same interval.

Function Inequality

A function inequality is a relationship between two functions where one is consistently greater than or equal to, or less than or equal to, another function over a certain interval. This can be express mathematically as \(0 \leq f(x) \leq g(x)\) for all \(x\) in a given range.

• This inequality speaks to the relationship between the values of \(f(x)\) and \(g(x)\), showing that \(f(x)\) is always less than or equal to \(g(x)\).
• In terms of definite integrals, this inequality allows us to directly conclude that the area under the curve of \(f(x)\) is less than or equal to the area under \(g(x)\) on the same interval.

This type of inequality is powerful, as it helps us understand how changes or differences in function values reflect in other calculations, such as areas and averages.
In the given exercise, understanding \(0 \leq f(x) \leq g(x)\) forms the basis upon which we can argue that the integral, and thus the average, of \(f(x)\) is less than or equal to the average of \(g(x)\).

Average Value of a Function

When we want to understand the overall behavior or effect of a function over a certain interval, finding its average value can be very illuminating. The average value of a function \( f(x)\) simply tells us what constant value could replace the function over the interval to have the same integral.
To compute this, use the formula \(\bar{y} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). This formula sums up all the function values over our interval and then divides by the length of the interval, essentially spreading its total effect evenly.
In our specific exercise case, the average value of \(f\) is calculated as \(\bar{y}_R = \frac{1}{1-0} \int_{0}^{1} f(x) \, dx\). Similarly, the average value of \(g\) is \(\bar{y}_S = \frac{1}{1-0} \int_{0}^{1} g(x) \, dx\). Because \(0 \leq f(x) \leq g(x)\) in the interval \([0,1]\), the average value \(\bar{y}_R\) of \(f(x)\) is also less than or equal to the average value \(\bar{y}_S\) of \(g(x)\).
Thus, understanding how to find and interpret average values of functions is essential for comparing overall function behaviors over specific intervals.