Question

Bounds on an integral Suppose \(f\) is continuous on \([a, b]\) with \(f^{\prime}(x)>0\) on the interval. It can be shown that $$ (b-a) f\left(\frac{a+b}{2}\right) \leq \int_{a}^{b} f(x) d x \leq(b-a) \frac{f(a)+f(b)}{2} $$ a. Assuming \(f\) is nonnegative on \([a, b],\) draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. Divide these inequalities by \((b-a)\) and interpret the resulting inequalities in terms of the average value of \(f\) on \([a, b]\)

Short answer

The inequalities can be interpreted in terms of the function's average value as: 1. The function value at the midpoint of the interval is less than or equal to the average value of the function on the entire interval. 2. The average value of the function on the entire interval is less than or equal to the average of the function values at its endpoints.

Step-by-step solution

a. Draw a figure illustrating the geometric meaning of the inequalities

To illustrate the geometric meanings of the inequalities, we can draw the graph of \(f(x)\) on the interval \([a, b]\). Since \(f'(x) > 0\) on the interval, \(f(x)\) is strictly increasing. Let's sketch the function \(f(x)\) on the interval \([a, b]\). The two inequalities give us the bounds on the area under \(f(x)\) from \(a\) to \(b\). 1. The left inequality states \((b-a)f\left(\frac{a+b}{2}\right) \leq \int_{a}^{b} f(x) dx\). We can draw the rectangle on the graph with base \([a, b]\) and height \(f\left(\frac{a+b}{2}\right)\). The area of this rectangle is the product of its base and height, which is \((b-a)f\left(\frac{a+b}{2}\right)\). The inequality implies that the area of this rectangle is less than or equal to the area under \(f(x)\) from \(a\) to \(b\). 2. The right inequality states \(\int_{a}^{b} f(x) dx \leq (b-a)\frac{f(a)+f(b)}{2}\). We can draw the trapezoid on the graph with bases \([a, b]\) and the vertices \([a, f(a)]\) and \([b, f(b)]\). The area of this trapezoid is the product of its height and the average of its bases' lengths, which is \((b-a)\frac{f(a)+f(b)}{2}\). The inequality implies that the area under \(f(x)\) from \(a\) to \(b\) is less than or equal to the area of this trapezoid. In conclusion, the inequalities show that the area under the strictly increasing continuous function \(f(x)\) on the interval \([a, b]\) is bounded by the area of a rectangle and a trapezoid, thereby providing a geometric meaning to these inequalities.

b. Interpret the resulting inequalities in terms of the average value of \(f\)

To interpret the inequalities in terms of the average value of \(f(x)\) on \([a, b]\), first divide the inequalities by \((b-a)\): $$ f\left(\frac{a+b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f(x) dx \leq \frac{f(a)+f(b)}{2}$$ The middle term, \(\frac{1}{b-a} \int_{a}^{b} f(x) dx\), represents the average value of \(f(x)\) on the interval \([a, b]\), denoted by \(f_{avg}\). The inequalities now have the following interpretation: 1. The left inequality, \(f\left(\frac{a+b}{2}\right) \leq f_{avg}\), implies that the function value at the midpoint of the interval is less than or equal to the average value of the function on the entire interval. 2. The right inequality, \(f_{avg} \leq \frac{f(a)+f(b)}{2}\), implies that the average value of the function on the entire interval is less than or equal to the average of the function values at its endpoints. These interpretations provide insights into the behavior and characteristics of strictly increasing continuous functions in relation to their average value.

Key concepts

Average Value of a Function

The average value of a function over an interval gives us a single number that summarizes the overall behavior of the function throughout that interval. When we say "average value," it is similar to how we might think of the average in daily life, but here it pertains to the values that the function takes. If you imagine plotting a curve of a function over an interval

• The average value represents a horizontal line whose total area of a rectangle under it is equal to an area under the curve.
• This involves integration, a process of calculating areas under curves.
• For a function \( f(x) \) that is defined on an interval \([a, b]\), the average value \( f_{avg} \) is given by the formula:

\[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]Breaking it down:

• "\(\int_{a}^{b} f(x) \, dx\)" computes the total area under the function \( f(x) \) from \( a \) to \( b \).
• By dividing this total area by the length of the interval "\((b-a)\)," you effectively distribute this "area" evenly, providing an "average height."

The average value gives us a way to understand what a function is doing across a range of inputs rather than just at specific points.

Continuous Functions

A continuous function is a critical concept in mathematics. A function \( f(x) \) is continuous on an interval if you can draw it without lifting your pencil. In practical terms, this means:

• There are no sudden jumps or breaks in the graph of \( f(x) \) within the specified interval.
• When we say a function is continuous on an interval \([a, b]\), every point \( c \) between \( a \) and \( b \) has a function value \( f(c) \), which approaches \( f \) values on either side as you get closer to \( c \).

Continuity is fundamental in calculus:

• It ensures the integrals exist between intervals like \([a, b]\).
• Continuous functions on closed intervals always attain maximum and minimum values, a key aspect in solving optimization problems.

Geometrically, continuous functions can fill the interval \([a, b]\) with their values without any gaps, which also implies the existence of integrals needed for calculating the average values.

Strictly Increasing Functions

A strictly increasing function is one where the function value grows as the input \( x \) grows. So, for any two numbers \( x_1 \) and \( x_2 \) in the interval, if \( x_1 < x_2 \), then \( f(x_1) < f(x_2) \).Key characteristics include:

• Strictly monotonic, meaning no two inputs give the same output; it's always rising.
• These functions guarantee uniqueness of solutions in problems, making them predictable.

In the context of integral bounds:

• The function is increasing from \( a \) to \( b \), so the lowest and highest values are at \( f(a) \) and \( f(b) \), respectively.
• This property helps in bounding the function's integrals, where the area under the curve is more easily contained between geometric shapes like rectangles and trapezoids.

Recognizing strictly increasing behavior provides insight into how functions behave in practical applications, facilitating predictions and evaluations across varying scenarios.