Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Decide whether the given statement is true or false. Then justify your answer. If \(\int_{a}^{b} f(x) d x \geq 0\), then \(f(x) \geq 0\) for all \(x\) in \([a, b] .\)

Step-by-Step Solution

Answer

False. The integral being non-negative doesn't imply \( f(x) \) is non-negative everywhere on \([a, b]\).

Step 1: Understand the Problem

We are given a condition that the definite integral of a function \( f(x) \) from \( a \) to \( b \) is greater than or equal to zero. We need to determine if this implies that \( f(x) \) is non-negative for every \( x \) in the interval \([a, b]\).

Step 2: Recall Properties of Definite Integrals

The definite integral \( \int_{a}^{b} f(x) \, dx \) represents the net area between the function \( f(x) \) and the x-axis over the interval \([a, b]\). If the integral is non-negative, it means the total area above the x-axis is greater than or equal to the area below the x-axis.

Step 3: Consider Counterexamples

To check the implication, consider a function \( f(x) \) that has areas both above and below the x-axis, but still results in a non-negative integral. A classic example is \( f(x) = x \) on \([-1, 1]\). Here, the integral is zero, but \( f(x) \) is negative for \( x \) in \(-1, 0)\).

Step 4: Conclusion on the Truth of the Statement

The statement is false. It's possible for \( f(x) \) to take negative values at some points in \([a, b]\) while still having an integral \( \int_{a}^{b} f(x) dx \geq 0 \). The integral only provides the net sum of areas above and below the x-axis.

Key Concepts

Definite Integral

In calculus, the concept of the definite integral is fundamental. It is used to compute the accumulation of quantities, such as areas under curves. Think of it as a tool that sums up infinitesimally small products of function values and widths over an interval.
The notation \( \int_{a}^{b} f(x) \, dx \) represents the definite integral of the function \( f(x) \) from \( x = a \) to \( x = b \). Here’s what you need to know about it:

• \( a \) and \( b \) are the limits of integration, marking the bounds of the interval along the x-axis.
• The definite integral measures the total accumulation of the quantity \( f(x) \) over \([a, b]\).
• It calculates a net sum that could include both areas above and below the x-axis.

The definite integral often provides insights into the overall pattern or trend of a function over a specified interval.

Net Area

The net area is a key concept to grasp when dealing with definite integrals. Essentially, it accounts for the geometry of a function related to the x-axis.
When we compute the integral \( \int_{a}^{b} f(x) \, dx \), it gives us the net area between the curve and the x-axis from \( x = a \) to \( x = b \). Here are some points to consider:

• Areas above the x-axis contribute positively to the integral.
• Areas below the x-axis contribute negatively.
• If there's more positive area than negative, the net area is positive; if it's balanced, the net area could be zero or even negative.

Therefore, having a non-negative definite integral doesn't necessarily mean the function is non-negative throughout the interval. The net effect can mask sections where the function dips below the x-axis.

Function Behavior

Understanding the behavior of functions is crucial when analyzing definite integrals. Functions can vary significantly within an interval even if their net effect seems positive.
While a non-negative definite integral suggests that the function predominantly resides above the x-axis, this is not a guarantee that it does so at every point.

• A function may be positive in some parts while being negative in others within the same interval.
• The function \( f(x) = x \) over the interval \([-1, 1]\) is a classic illustration where the integral is zero but the function itself varies through positive and negative values.
• This disparity arises because integration focuses on the overall trend, balancing out periods of positive contribution with negative ones.

Grasping this dynamic helps clarify why a net positive definite integral doesn't imply all function values are non-negative across the interval.