Problem 131 - Calculus Volume 1

Question

Suppose \(f\) and \(g\) are continuous functions such that \(\int_{c}^{d} f(t) d t \leq \int_{c}^{d} g(t) d t\) for every subinterval \([c, d]\) of \([a, b] .\) Explain why \(f(x) \leq g(x)\) for all values of \(x .\)

Step-by-Step Solution

Answer

Since integrals of \(f\) are smaller than those of \(g\) over all subintervals, \(f(x) \leq g(x)\) for all \(x\).

Step 1: Understand the Given Condition

The problem states that for any subinterval \([c, d]\) within \([a, b]\), the integral of \(f(t)\) from \(c\) to \(d\) is less than or equal to the integral of \(g(t)\) over the same interval. Mathematically, this is expressed as \(\int_{c}^{d} f(t) dt \leq \int_{c}^{d} g(t) dt\). This condition holds for all possible subintervals within \([a, b]\).

Step 2: Apply the Fundamental Theorem of Calculus

Since \(f\) and \(g\) are continuous functions on \([a, b]\), by the Fundamental Theorem of Calculus, the integral \(\int_{c}^{d} f(t) dt\) is equivalent to \(F(d) - F(c)\), where \(F\) is an antiderivative of \(f\). Similarly, \(\int_{c}^{d} g(t) dt = G(d) - G(c)\), where \(G\) is an antiderivative of \(g\). Under the given condition, we have \(F(d) - F(c) \leq G(d) - G(c)\) for every subinterval \([c, d]\).

Step 3: Deduce the Inequality of Derivatives

By the Mean Value Theorem for integrals, for any point \(x\) within the interval \([c, d]\), the average value of \(f(t)\) and \(g(t)\) over \([c, d]\) can be expressed as \(f(x) = \frac{F(d) - F(c)}{d - c}\) and \(g(x) = \frac{G(d) - G(c)}{d - c}\). Since \(F(d) - F(c) \leq G(d) - G(c)\), this implies \(f(x) \leq g(x)\) for every choice of \(x\) in \([a, b]\).

Step 4: Conclusion

Thus, due to the continuity of \(f\) and \(g\) on \([a, b]\) and the given integral inequality for all subintervals \([c, d]\), it is concluded that \(f(x) \leq g(x)\) holds for all \(x\) in the interval \([a, b]\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key concept that connects differentiation and integration, two of the central ideas in calculus. It has two main parts, and understanding these will help you see how integration and differentiation are inverse processes.

The first part of the theorem states that if you have a continuous function, say \( f(x) \), and you integrate it over an interval \([a, b]\), the result can be found using an antiderivative \( F(x) \). Mathematically, this is expressed as \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). It tells us that finding the definite integral of \( f(x) \) between \( a \) and \( b \) means calculating the change in the antiderivative \( F(x) \) over that interval.

The second part of the theorem shows that if \( F(x) \) is defined as \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative of \( F(x) \) with respect to \( x \) is \( f(x) \). This illustrates how differentiation reverses the process of integration. In the context of the original exercise, knowing that \( F(d) - F(c) \) equals the integral of \( f(t) \) over \([c, d]\), you can comprehend how comparing these integral results leads to an understanding of the function values \( f(x) \) and \( g(x) \). By identifying the inequalities of their integral sums, you deduce the pointwise inequality \( f(x) \leq g(x) \).

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals provides a way to find the "average" value of a function over an interval. Imagine you have a continuous function \( f(x) \) on \([a, b]\). According to this theorem, there exists at least one point \( c \) in \([a, b]\) where the value of the function equals the average value of the function on that interval.

Mathematically, the Mean Value Theorem for Integrals states that \( f(c) = \frac{1}{b-a}\int_{a}^{b} f(x) \, dx \). This represents an average height of \( f(x) \) over \([a, b]\).

In solving problems like the original exercise, this theorem helps to relate the integrals of \( f(x) \) and \( g(x) \) to their values at specific points. If \( \int_{c}^{d} f(t) \, dt \leq \int_{c}^{d} g(t) \, dt \), it implies that the average value of \( f(t) \) over \([c, d]\) is less than or equal to that of \( g(t) \). Via the theorem, you can pinpoint that \( f(x) \leq g(x) \) for every point \( x \) in \([a, b]\) as found in the original solution.

Continuous Functions

Continuous functions are the building blocks of calculus, and they have special properties that make them particularly useful in analysis. A function \( f(x) \) is considered continuous on an interval if, loosely speaking, you can draw it without lifting your pen from the paper.

Formalizing this idea, \( f(x) \) is continuous at a point \( c \) if the limit as \( x \) approaches \( c \) matches the function value at \( c \): \( \lim_{x\to c} f(x) = f(c) \). When a function is continuous on an interval \([a, b]\), it's uniform in behavior, and there are no sudden jumps or breaks.

Continuous functions allow us to apply theorems like the Fundamental Theorem of Calculus and the Mean Value Theorem because they guarantee that the functions are well-behaved and follow predictable patterns. In the original example, having \( f(x) \) and \( g(x) \) as continuous ensures their integrals are valid over any subinterval within \([a, b]\). It’s this property of continuity that helps to establish and apply integral inequalities and deduce further characteristics of \( f(x) \) and \( g(x) \), validating conclusions like \( f(x) \leq g(x) \) pointwise across the interval.