Problem 58
Question
Decide whether the given statement is true or false. Then justify your answer. If \(f(x) \geq 0\) and \(\int_{a}^{b} f(x) d x=0,\) then \(f(x)=0\) for all \(x\) in \([a, b]\).
Step-by-Step Solution
Verified Answer
The statement is true. For a non-negative function, if the integral is zero, the function must be zero everywhere on the interval.
1Step 1: Understanding the Statement
We are asked to determine whether the statement "If \(f(x) \geq 0\) and \(\int_{a}^{b} f(x) \, dx=0,\) then \(f(x)=0\) for all \(x\) in \([a, b]\)" is true or false. First, understand that \(f(x)\) is non-negative on the interval \([a, b]\). The integral of a non-negative function being equal to zero has specific implications.
2Step 2: Analyzing the Integral Property
An integral \(\int_{a}^{b} f(x) \, dx=0\) indicates the net area between the function \(f(x)\) and the x-axis over the interval \([a, b]\) is zero. For a non-negative function, this can only happen if there is no area, implying \(f(x) = 0\) for all \(x\) in \([a, b]\).
3Step 3: Drawing Conclusion from Non-negativity
Since \(f(x) \geq 0\) on \([a, b]\), the only way the integral can equal zero is if \(f(x) = 0\) at every point in the interval. If \(f(x)\) were positive even at a single point within the interval, the integral would be positive, contradicting the given \(\int_{a}^{b} f(x) \, dx=0\).
4Step 4: Final Verdict on the Statement
Based on the analysis, the statement is true. For \(\int_{a}^{b} f(x) \, dx\) to be zero while \(f(x)\) is non-negative, \(f(x)\) must indeed be zero at every point within the interval \([a, b]\).
Key Concepts
Non-negative FunctionIntegral PropertyNet Area
Non-negative Function
A non-negative function is a function where all of its outputs are equal to or greater than zero over a specified interval. This means that for any point \(x\) in the interval \([a, b]\), the value of the function \(f(x)\) will always satisfy \(f(x) \geq 0\).
Non-negative functions have characteristics that make them particularly interesting when dealing with definite integrals. When you integrate a non-negative function over an interval, you are essentially calculating the area under the curve from the x-axis to the top of the function, above the interval \([a, b]\). This property will be helpful in understanding integrals and their implications.
In relation to definite integration, non-negative functions cannot produce negative net areas because there isn't a point at which their values drop below zero. This characteristic is crucial when analyzing the implications of integrals equating to zero.
Non-negative functions have characteristics that make them particularly interesting when dealing with definite integrals. When you integrate a non-negative function over an interval, you are essentially calculating the area under the curve from the x-axis to the top of the function, above the interval \([a, b]\). This property will be helpful in understanding integrals and their implications.
In relation to definite integration, non-negative functions cannot produce negative net areas because there isn't a point at which their values drop below zero. This characteristic is crucial when analyzing the implications of integrals equating to zero.
Integral Property
The integral of a function from \(a\) to \(b\) is a fundamental concept in calculus known as the definite integral. It represents the signed area under the curve of the function across the interval \([a, b]\), and is expressed as \(\int_{a}^{b} f(x) \, dx\).
For a non-negative function, this integral directly characterizes the total "net area" above the x-axis that the function covers over the given interval.
When this integral equals zero, as the case in our example, it conveys an important message: if the function \(f(x)\) is non-negative throughout \([a, b]\), the only possible scenario for a zero integral is if \(f(x) = 0\) for every point within that interval.
To put this into perspective, consider that if the function had any points where it was greater than zero, the integral would reflect a positive result due to the areas that contribute above the x-axis. This is why the integral property is key in understanding the distinction between non-positive and zero-valued regions.
For a non-negative function, this integral directly characterizes the total "net area" above the x-axis that the function covers over the given interval.
When this integral equals zero, as the case in our example, it conveys an important message: if the function \(f(x)\) is non-negative throughout \([a, b]\), the only possible scenario for a zero integral is if \(f(x) = 0\) for every point within that interval.
To put this into perspective, consider that if the function had any points where it was greater than zero, the integral would reflect a positive result due to the areas that contribute above the x-axis. This is why the integral property is key in understanding the distinction between non-positive and zero-valued regions.
Net Area
The concept of net area plays a vital role in understanding the outcome of definite integrals, particularly when dealing with non-negative functions. Net area refers to the total area that lies between the function and the x-axis over a specific interval.
For a non-negative function, the net area is essentially a measure of how much space the function occupies above the x-axis. It ensures a clear understanding that the calculated area through integration does not include any 'negative' contributions (since the function doesn't dip below the x-axis).
Thus, when the net area is zero for a non-negative function, it visually and mathematically implies that the function does not rise above the x-axis at all. In simplest terms, the function \(f(x)\) stays pinned exactly on the x-axis throughout every single point of the interval.
This understanding helps reinforce why in such cases, the function must be zero at all points within the interval to achieve a zero net area. The integrity of this principle is what validates the given statement as true, emphasizing the relationship between function properties and their integral results.
For a non-negative function, the net area is essentially a measure of how much space the function occupies above the x-axis. It ensures a clear understanding that the calculated area through integration does not include any 'negative' contributions (since the function doesn't dip below the x-axis).
Thus, when the net area is zero for a non-negative function, it visually and mathematically implies that the function does not rise above the x-axis at all. In simplest terms, the function \(f(x)\) stays pinned exactly on the x-axis throughout every single point of the interval.
This understanding helps reinforce why in such cases, the function must be zero at all points within the interval to achieve a zero net area. The integrity of this principle is what validates the given statement as true, emphasizing the relationship between function properties and their integral results.
Other exercises in this chapter
Problem 57
Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{1} x \cos ^{3}\left(x^{2}\right) \sin \left(x^{2}\right) d x $
View solution Problem 58
Find the area of the region under the curve \(y=f(x)\) over the interval \([a, b] .\) To do this, divide the interval \([a, b]\) into \(n\) equal subintervals,
View solution Problem 58
Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-\pi / 2}^{\pi / 2} x^{2} \sin ^{2}\left(x^{3}\right) \cos \left(x
View solution Problem 59
If \(\int_{a}^{b} f(x) d x>\int_{a}^{b} g(x) d x,\) then \(\quad \int_{a}^{b}[f(x)-g(x)] d x>0\)
View solution