Problem 54 - Calculus: Early Transcendentals

Question

Suppose \( f \) has absolute minimum value \( m \) and absolute maximum value \( M \). Between what two values must \( \displaystyle \int^2_0 f(x) \,dx \) lie? Which property of integrals allows you to make your conclusion?

Step-by-Step Solution

Answer

The integral lies between \( 2m \) and \( 2M \) by the properties of definite integrals.

Step 1: Understanding the Function's Minimum and Maximum

The function \( f \) has an absolute minimum value \( m \) and an absolute maximum value \( M \). This means that for all \( x \) in the interval \([0, 2]\), the function satisfies \( m \leq f(x) \leq M \).

Step 2: Applying the Properties of Definite Integrals

The property of definite integrals tells us that if a function \( f(x) \) is bounded by \( m \leq f(x) \leq M \) over an interval \([a, b]\), then the integral of \( f \) from \( a \) to \( b \) is similarly bounded:\[m(b - a) \leq \int_{a}^{b} f(x) \, dx \leq M(b - a).\]

Step 3: Calculate the Bounds Using the Interval Length

For the given problem, the interval \([0, 2]\) has a length \( b - a = 2 - 0 = 2 \). Thus, the integral bounds are calculated as follows:\[2m \leq \int_{0}^{2} f(x) \, dx \leq 2M.\]

Step 4: Conclusion

Therefore, the integral \( \int_{0}^{2} f(x) \, dx \) is guaranteed to lie between \( 2m \) and \( 2M \). This conclusion is reached using the properties of definite integrals that allow us to estimate the bounds based on the minimum and maximum values of the function.

Key Concepts

Absolute Minimum and Maximum

An absolute minimum of a function is the smallest value that the function reaches over its entire domain, while the absolute maximum is the largest value. When we describe a function as having an absolute minimum or maximum, it means that these values are the lowest and highest, respectively, that the function attains everywhere on its given interval.

For our exercise, we consider a function \( f(x) \) defined on the interval \([0, 2]\) with an absolute minimum value \( m \) and an absolute maximum value \( M \). Therefore, every value that the function takes on within this interval is somewhere between \( m \) and \( M \). This understanding is crucial because it sets the stage for determining the possible values of the integral over this interval.

Bounded Functions

A function is said to be bounded on an interval when there exist real numbers that form a limit to the function's values over that interval. Specifically for an interval \([a, b]\), a function \( f(x) \) is bounded if there are numbers \( m \) and \( M \) such that \( m \leq f(x) \leq M \) for every \( x \) within \([a, b]\).

• The lower bound \( m \) represents the lowest value \( f \) can take.
• The upper bound \( M \) represents the highest value \( f \) can take.

In our scenario, the function \( f(x) \) is bounded by the values \( m \) and \( M \) across the interval \([0, 2]\). This means that no matter what, the function's output will not sink below \( m \) or rise above \( M \). This bound is critical when using definite integrals to define a range of possible outcomes, which is achieved through the use of integral properties.

Integrals Properties

Definite integrals represent the signed area under a curve between two limits on a graph. A useful property is that the integral of a bounded function over an interval has its own boundaries based on the bounds of the function. For any function \( f(x) \) that is bounded by \( m \leq f(x) \leq M \) over an interval \([a, b]\), the property of integrals states:

• \( m(b - a) \leq \int_{a}^{b} f(x) \, dx \leq M(b - a) \)

This is because the "area" that these bounds cover is the product of the length of the interval \((b-a)\) and the bounds \( m \) and \( M \). So, the minimum possible value of the integral is \( m(b-a) \), and the maximum is \( M(b-a) \).

In our exercise, the interval is from 0 to 2, giving a length of 2. Therefore:

• The integral overall will lie between \( 2m \) and \( 2M \).

This use of integral properties allows us to precisely estimate where the definite integral should fall, ensuring that we understand how integrals can be bounded by the function's minimum and maximum values.