Problem 84 - Thomas Calculus

Question

Show that if \(f\) is continuous, then $$\int_{0}^{1} f(x) d x=\int_{0}^{1} f(1-x) d x$$

Step-by-Step Solution

Answer

By substitution, both integrals are equal.

Step 1: Express the Problem Statement

We need to prove that for a continuous function \( f \), the integrals \( \int_{0}^{1} f(x) \, dx \) and \( \int_{0}^{1} f(1-x) \, dx \) are equal.

Step 2: Use the Substitution Method

Consider the substitution \( u = 1 - x \), which implies that \( du = -dx \). Under this substitution, when \( x = 0 \), \( u = 1 \), and when \( x = 1 \), \( u = 0 \). This changes our integration limits.

Step 3: Perform the Substitution in the Integral

Substitute in the integral \( \int_{0}^{1} f(1-x) \, dx \). It becomes:\[ \int_{1}^{0} f(u) (-du) \]which simplifies to:\[ \int_{0}^{1} f(u) \, du \]

Step 4: Compare the Two Integrals

Now we have both integrals in terms of \( f \) over the same interval:- \( \int_{0}^{1} f(x) \, dx \)- \( \int_{0}^{1} f(u) \, du \)Since both represent integration of the continuous function \( f \) over \([0, 1] \), they are equal.

Key Concepts

Continuous Functions

In the realm of calculus, a continuous function is a fundamental concept. Put simply, a function is considered continuous if there are no interruptions in its graph. Imagine drawing the function on a piece of paper without lifting your pen, that's continuity. Continuous functions are vital because they ensure stability in the behavior of functions and the results they generate.

They feature certain properties that make them essential for calculus, particularly integral calculus:

• The Intermediate Value Theorem states that if a function is continuous on an interval, any value between the function's extremes in that interval is also assumed. This property supports the finding of roots or zeroes efficiently.
• Limits are easily dealt with in continuous functions, allowing seamless computation of function values at particular points.
• Continuous functions on a closed interval are bounded and attain their maximum and minimum values, a property useful for solving integration problems.

Understanding that a function is continuous allows us to use certain techniques in integration, such as the substitution method, with confidence that the fundamental behavior of the function won't lead to unexpected results.

Substitution Method in Integration

The substitution method in integration is a powerful tool that simplifies the integration process. It is analogous to the "reverse" of the chain rule used in differentiation. The goal of this method is to transform a difficult integral into a simpler one that is easier to solve.

Here's how it works:

• A substitution is chosen, often denoted as \( u \), to replace a part of the integral. This choice is usually guided by a part of the integral that undergoes differentiation seamlessly.
• The differential \( dx \) is replaced accordingly with \( du \), requiring accounting for any constants or changes in limits when dealing with definite integrals.
• Any integral boundaries are transformed according to the new variable. For instance, if the original integral has limits from \( a \) to \( b \), the limits become \( u(a) \) and \( u(b) \).
• The function is re-evaluated in terms of \( u \), simplifying the integration process.

In this exercise, the substitution \( u = 1-x \) helped to prove the equality of two integrals. The technique allowed for the manipulation of limits and showcased how simple transformations could maintain equality in a continuous function's integral.

Definite Integrals

Definite integrals represent the accumulation of quantities, which could be areas under a curve, over a specific interval. Unlike indefinite integrals, definite integrals compute the net area and provide an actual number rather than a family of functions.

Key components:

• The limits of integration, shown as the lower and upper limits, define the interval across which the function is integrated.
• Continuous functions on closed intervals are crucial because they assure the existence of a definite integral, thanks to their predictable nature.
• The Fundamental Theorem of Calculus links differentiation with integration, stating that if a function is continuous over an interval, then its integral can be computed using its antiderivative.

In this problem, definite integrals of a continuous function over the interval \([0, 1]\) were shown to be the same, highlighting an interesting aspect of integration. The property that \( \int_{0}^{1} f(x) \, dx = \int_{0}^{1} f(1-x) \, dx \) is a fine example of symmetry and stability in continuous functions and their integrals.