Question

Decide whether the given statement is true or false. Then justify your answer. $$\text { If } \begin{aligned} \int_{a}^{b} f(x) d x &>\int_{a}^{b} g(x) d x, \text { then } \\ \int_{a}^{b}[f(x)-g(x)] d x>0 \end{aligned}$$

Short answer

True; the integrals difference confirms \( \int_{a}^{b} [f(x)-g(x)] dx > 0 \).

Step-by-step solution

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.

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.

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.

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

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.

• 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.

By mastering the basics of calculus, such as how to compute integrals, you open the door to solving many real-world situations involving change and motion.

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:

• 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: \[ \int_{a}^{b} c f(x) \, dx = c \int_{a}^{b} f(x) \, dx \]

This property greatly aids in breaking down complex expressions into more manageable ones.

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.

• 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.

In the original problem, this concept was applied to determine not just how the integrals of two functions compared, but also how their difference, when expressed as an integral itself, must behave.
This knowledge is particularly useful when trying to establish bounds or limits for solutions in real-world problems.