Problem 58
Question
Decide whether the given statement is true or false. Then justify your answer. If \(f(x) \geq 0\) and \(\int_{a} f(x) d x=0\), then \(f(x)=0\) for all \(x\) in \([a, b]\).
Step-by-Step Solution
Verified Answer
The statement is true; if \(\int_a^b f(x) \, dx = 0\) and \(f(x) \geq 0\), then \(f(x) = 0\) for all \(x\) in \([a, b]\).
1Step 1: Understand the Statement
The problem asks whether a non-negative function \(f(x)\) (where \(f(x) \geq 0\) for all \(x\) in the interval \([a, b]\)) that has an integral \(\int_a^b f(x) \, dx = 0\), must be zero for all \(x\) in \([a, b]\).
2Step 2: Recall the Properties of Integration
If a function \(f(x)\) is non-negative over an interval \([a, b]\), the value of the definite integral \(\int_{a}^{b} f(x) \, dx\) quantifies the 'area under the curve' from \(a\) to \(b\). The integral equals zero when the area is zero.
3Step 3: Analyze the Condition \(\int_{a}^{b} f(x) \, dx = 0\)
Since the integral of a non-negative function is the total area under the curve, if this integral equals zero, it implies that \(f(x)\) must have no 'height' at any point in \([a, b]\) that contributes to the area, i.e., \(f(x)\) must be zero for all \(x\) in \([a, b]\).
4Step 4: Conclude Based on Analysis
If \(f(x)\) were positive at any point on \([a, b]\), the integral would be positive due to the non-negative nature of the function. Thus, the condition \(\int_{a}^{b} f(x) \, dx = 0\) logically and mathematically forces \(f(x)\) to be zero at all points on \([a, b]\).
5Step 5: State the Conclusion
The statement is true because a non-negative function that integrates to zero over a closed interval must be zero at every point of that interval.
Key Concepts
Non-Negative FunctionsDefinite IntegralsFundamental Theorem of Calculus
Non-Negative Functions
A non-negative function is one that never dips below the horizontal axis, meaning the function's value is zero or positive for all inputs in its domain. This is represented mathematically as \(f(x) \geq 0\) for every \(x\) in the interval.
In cases where a question states that \(f(x)\) is non-negative, it implies all current and potential calculations maintain this baseline of non-negativity. This foundational principle ensures that our further calculations or assumptions start on solid ground.
- These functions are essential in calculus, as they define a region that is either at or above the x-axis.
- Since they never have \(y\) values less than zero, when graphed, they create regions we can calculate areas for using integrals.
- Understanding this ensures clarity when discussing areas and integrals, since negative areas might indicate potential misinterpretations.
In cases where a question states that \(f(x)\) is non-negative, it implies all current and potential calculations maintain this baseline of non-negativity. This foundational principle ensures that our further calculations or assumptions start on solid ground.
Definite Integrals
In calculus, definite integrals are powerful tools used to calculate the total 'accumulated quantity' for a function over an interval \([a, b]\).
If the function is non-negative throughout the interval, a zero value for the integral signifies no area under the curve at all, signaling the function is constantly zero over that range.
- This 'accumulated quantity' is often visualized as the area under the curve of a function, between the two points \(a\) and \(b\).
- The notation \(\int_{a}^{b} f(x) \, dx \) specifies the function \(f(x)\) and the range \([a, b]\) over which this area accumulates.
- The calculated net area can be positive, negative, or zero, depending on the function's position relative to the x-axis.
If the function is non-negative throughout the interval, a zero value for the integral signifies no area under the curve at all, signaling the function is constantly zero over that range.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus interconnects the concept of differentiation and integration, the two main activities in calculus. It provides a way to evaluate definite integrals and relate them directly to antiderivatives.
Understanding this theorem is crucial for solving integrals efficiently, as it directly links the rules of differentiation to those of integration, creating a seamless flow in calculus problem solving.
- The theorem states that if \(F(x)\) is an antiderivative of a function \(f(x)\), then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\).
- In simple terms, in practical applications, once you know the antiderivative, you can easily calculate the integral over an interval by evaluating the difference at the endpoints.
- This theorem simplifies finding definite integrals, bypassing the need for Riemann sums or complex limit processes.
Understanding this theorem is crucial for solving integrals efficiently, as it directly links the rules of differentiation to those of integration, creating a seamless flow in calculus problem solving.
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Problem 58
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