Problem 59
Question
Decide whether the given statement is true or false. Then justify your answer. $$\text { If } \begin{aligned} \int_{a}^{b} f(x) d x &>\int_{a}^{b} g(x) d x, \text { then } \\ \int_{a}^{b}[f(x)-g(x)] d x>0 \end{aligned}$$
Step-by-Step Solution
Verified Answer
True; the integrals difference confirms \( \int_{a}^{b} [f(x)-g(x)] dx > 0 \).
1Step 1: Understanding the Exercise
We are given two integrals, \( \int_{a}^{b} f(x) \, dx \) and \( \int_{a}^{b} g(x) \, dx \), with the inequality \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \). We need to determine whether this implies \( \int_{a}^{b} [f(x)-g(x)] \, dx > 0 \) and justify the answer.
2Step 2: Applying Linearity of Integrals
The linearity of integrals tells us that \( \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \). This property allows us to separate the subtraction inside the integrals.
3Step 3: Analyzing the Given Inequality
We know from the problem that \( \int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx \). This directly implies that their difference, \( \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \), is greater than zero.
4Step 4: Conclusion Based on Analysis
The expression \( \int_{a}^{b} [f(x) - g(x)] \, dx > 0 \) is equivalent to \( \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx > 0 \). Since we have determined this from the given inequality, the statement is indeed true.
Key Concepts
CalculusLinearity of IntegralsIntegral Comparison
Calculus
Calculus is a branch of mathematics that studies how things change and accumulate. It is primarily divided into two parts: differentiation and integration.
Differentiation focuses on finding rates of change or slopes of curves. Integration, on the other hand, involves summing up small parts to determine a whole quantity. This can be thought of as finding the area under a curve. Both processes are interconnected, with integration often viewed as the reverse of differentiation.
Understanding calculus is crucial because it allows us to model and solve complex problems in various fields such as physics, engineering, and economics.
Differentiation focuses on finding rates of change or slopes of curves. Integration, on the other hand, involves summing up small parts to determine a whole quantity. This can be thought of as finding the area under a curve. Both processes are interconnected, with integration often viewed as the reverse of differentiation.
Understanding calculus is crucial because it allows us to model and solve complex problems in various fields such as physics, engineering, and economics.
- Calculus helps us understand how quantities grow or shrink.
- It provides tools for finding volumes, areas, and lengths.
- It plays a pivotal role in optimizing functions to get the best results.
Linearity of Integrals
The linearity of integrals is a fundamental property in calculus that simplifies the computation of integrals by allowing one to manage sums and scalar multiples of functions easily.
This property states:
In the context of the original problem, linearity was used to simplify the expression \(\int_{a}^{b} [f(x) - g(x)] \, dx\) into more familiar terms \(\int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx\), thus making it easier to see how the given inequality applies.
Recognizing and applying the linearity of integrals can make solving calculus problems more intuitive and less error-prone.
This property states:
- If you have two functions, say \(f(x)\) and \(g(x)\), then the integral of their sum is the sum of their integrals: \[ \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \]
- For any constant \(c\), the integral of \(c\) times a function is \(c\) times the integral of the function: \[ \int_{a}^{b} c f(x) \, dx = c \int_{a}^{b} f(x) \, dx \]
In the context of the original problem, linearity was used to simplify the expression \(\int_{a}^{b} [f(x) - g(x)] \, dx\) into more familiar terms \(\int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx\), thus making it easier to see how the given inequality applies.
Recognizing and applying the linearity of integrals can make solving calculus problems more intuitive and less error-prone.
Integral Comparison
The concept of integral comparison is about comparing the sizes or values of different integrals.
This involves looking at two functions, \(f(x)\) and \(g(x)\), over the same interval \([a, b]\) to see which function accumulates more value between these points.
If you know that \(\int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx\), it means that the total "area under the curve" of \(f(x)\) is greater than that of \(g(x)\) over the specified interval.
This is useful in understanding how different functions behave comparatively over an interval.
This knowledge is particularly useful when trying to establish bounds or limits for solutions in real-world problems.
This involves looking at two functions, \(f(x)\) and \(g(x)\), over the same interval \([a, b]\) to see which function accumulates more value between these points.
If you know that \(\int_{a}^{b} f(x) \, dx > \int_{a}^{b} g(x) \, dx\), it means that the total "area under the curve" of \(f(x)\) is greater than that of \(g(x)\) over the specified interval.
This is useful in understanding how different functions behave comparatively over an interval.
- Integral comparison helps us assess how functions grow or shrink in relation to one another.
- It is crucial in various applications, including physics, where it is important to know how one quantity measures up against another over time or space.
This knowledge is particularly useful when trying to establish bounds or limits for solutions in real-world problems.
Other exercises in this chapter
Problem 58
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