Question

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

Short answer

False. The integral can be zero even if \( f(x) \neq 0 \) in some areas of the interval due to positive and negative parts canceling out.

Step-by-step solution

Understanding the given statement

We are given that the integral of a function \( f(x) \) from \( a \) to \( b \) is zero, i.e., \( \int_{a}^{b} f(x) \, dx = 0 \). The statement asserts that if this is true, then \( f(x) \) must be zero for all \( x \) in the interval \([a, b]\). We need to decide if this is true or false.

Consider the properties of integrals

The integral \( \int_{a}^{b} f(x) \, dx \) being zero implies the net area between the function \( f(x) \) and the \( x \)-axis over the interval \([a, b]\) is zero. This does not require \( f(x) \) to be zero everywhere on \([a, b]\). Instead, \( f(x) \) could be positive on some parts and negative on others, leading to a net sum of zero.

Counterexample check

To further evaluate the statement, consider a counterexample. Suppose \( f(x) \) is defined as follows within this interval: \( f(x) = \begin{cases} 1, & \text{for } x \in [a, c) \ -1, & \text{for } x \in [c, b] \end{cases} \), where \( c \) is any number between \( a \) and \( b \). For this function, \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} 1 \, dx + \int_{c}^{b} (-1) \, dx = (c-a) - (b-c) = 0 \), yet \( f(x) eq 0 \) on \([a, b]\).

Conclusion

Based on the counterexample and the integral property, the given statement is false. \( f(x) \) does not need to be zero for all \( x \in [a, b] \) for its integral over \([a, b]\) to be zero.

Key concepts

Definite Integrals

In calculus, one of the fundamental concepts you'll encounter is the **definite integral**. It represents the accumulation of quantities, and it can be visualized as the area under a curve between two points. Given by the notation \(\int_{a}^{b} f(x) \, dx\), the definite integral calculates the net area between the function \(f(x)\) and the \(x\)-axis from \(a\) to \(b\).
For students new to this concept, it's important to note that the definite integral takes into account areas above the \(x\)-axis as positive and below the \(x\)-axis as negative. This means that the "total" area can cancel out, leaving a "net" area that could be zero, even if the function isn't zero along the interval.
This property becomes pivotal in exercises where you're asked to conclude something about the function \(f(x)\) based on the integral's value.

Properties of Integrals

Understanding the properties of integrals is crucial for solving many calculus problems. One key property is **linearity**, which states that \(\int (c_1f(x) + c_2g(x)) \, dx = c_1\int f(x) \, dx + c_2\int g(x) \, dx\), where \(c_1\) and \(c_2\) are constants.

Another important property is that of **additivity** with intervals. For instance, \(\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\), which reflects how integrals over contiguous intervals can be added or subtracted. Also, the **fundamental theorem of calculus** connects the process of integration with differentiation, creating a bridge between these two primary concepts in calculus.
These properties highlight that an integral can be zero even if the function is non-zero over its domain, since different sections of the function may balance each other out.

Net Area

The term **net area** in calculus refers to the concept that areas under a curve can cancel each other out. When you compute \(\int_{a}^{b} f(x) \, dx\), you're essentially summing over all positive and negative areas formed by the function \(f(x)\) over the interval \([a, b]\).

The net area illustrates why a definite integral's result could be zero, even with non-zero values of \(f(x)\). In scenarios where a function switches from positive to negative values (or vice versa), the areas can cancel out, leading to this zero net area.

As seen in the problem's step by step explanation, a function can have sections above and below the \(x\)-axis that together sum to zero, demonstrating there is no "absolute" area. The concept of net area underscores the importance of considering the entire behavior of the function across the interval, rather than focusing on isolated values.