Problem 76 - Calculus Early Transcendentals

Question

Show that if \(f\) and \(g\) have continuous second derivatives and \(f(0)=f(1)=g(0)=g(1)=0,\) then $$ \int_{0}^{1} f^{\prime \prime}(x) g(x) d x=\int_{0}^{1} f(x) g^{\prime \prime}(x) d x $$

Step-by-Step Solution

Answer

Question: Show that \(\int_{0}^{1} f^{\prime \prime}(x) g(x) dx = \int_{0}^{1} f(x) g^{\prime \prime}(x) dx\) given that \(f(1) = f(0) = g(1) = g(0) = 0\). Solution: By applying integration by parts twice on the left-hand side of the equation and using the given boundary conditions, we showed that the integral of the product of the second derivatives of two functions \(f\) and \(g\) equals the integral of the product of the functions and their second derivatives. Therefore, \(\int_{0}^{1} f^{\prime \prime}(x) g(x) dx = \int_{0}^{1} f(x) g^{\prime \prime}(x) dx\).

Step 1: Apply integration by parts to the LHS expression

Let us first give names to the two integrals we need to deal with. Let \(I_1 = \int_{0}^{1} f^{\prime \prime}(x) g(x) dx\) and \(I_2 = \int_{0}^{1} f(x) g^{\prime \prime}(x) dx\). We will demonstrate that \(I_1 = I_2\). Start with \(I_1\) and apply integration by parts with \(u = g(x)\) and \(v' = f''(x)\). According to integration by parts, \(\int u dv = uv - \int v du\). So, \(I_1 = g(x)f'(x) \bigg |_{0}^1 - \int_{0}^{1} g'(x)f'(x) dx\). Notice that the boundary term vanishes because \(f(1) = f(0) = g(1) = g(0) = 0\). Thus, \(I_1 = - \int_{0}^{1} g'(x)f'(x) dx\).

Step 2: Apply integration by parts again to the LHS expression

Now, apply integration by parts another time with \(u = g'(x)\) and \(v' = f'(x)\). \(I_1 = - [g'(x)f(x) \bigg |_{0}^1 - \int_{0}^{1} g''(x)f(x) dx]\) Once again, the boundary term vanishes due to the conditions given. So, we are left with: \(I_1 = \int_{0}^{1} f(x) g^{\prime \prime}(x) d x\)

Step 3: Conclude that \(I_1= I_2\)

Since we have shown that \(I_1 = \int_{0}^{1} f(x) g^{\prime \prime}(x) d x\), we can conclude that \(I_1 = I_2\), and the given expression is true.

Key Concepts

Continuous Second Derivatives

One fundamental concept in calculus is the idea of continuous second derivatives. This concept is essential when dealing with problems that involve curvature, smoothness, and the behavior of functions over an interval. A function has a continuous second derivative if its second derivative exists and is a continuous function. This means that not only does the function curve smoothly without any sharp turns or corners (think of the path a car could easily drive on), but also that its rate of change of the slope is consistent throughout its domain.

When functions have continuous second derivatives, they are often nicer to work with in calculus, particularly with techniques such as integration by parts, because they follow well-behaved patterns. Moreover, the existence of continuous second derivatives ensures that functions don't have unpredictable behavior, which could complicate integration and other analyses.

Integrating Functions with Continuous Second Derivatives

In the exercise provided, both functions f and g have continuous second derivatives, which allows us to apply integration by parts confidently. The assurance that no discontinuities or undefined regions will disrupt the evaluation of the integrals is what makes integration by parts viable here and leads us to conclude that two seemingly different integrals are, in fact, equal.

Integral Calculus

Integral calculus is a branch of mathematics focusing on the accumulation of quantities and the areas under and between curves. Essentially, integration is the process of finding the integral of a function, which can represent area, volume, displacement, and more, depending on the context. The definite integral, which we encounter in this exercise, is concerned with the net accumulation of a quantity on a closed interval, like the area under a curve from one point to another.

Integration by parts is a technique used within integral calculus and is based on the product rule for differentiation. It allows us to transform the integral of a product of functions into simpler integrals that we can more easily evaluate. In the solved exercise, integration by parts is used as a strategic tool to demonstrate a relationship between two integrals involving second derivatives.

Applying Integration by Parts

One of the key factors of integration by parts is choosing which function to differentiate and which to integrate. This choice influences the complexity of the subsequent integrals. With this exercise, integration by parts simplifies the initial complex-looking integral into a format that utilizes the given boundary conditions, leading to a much simpler form.

Calculus Education

Calculus education involves teaching and learning the fundamental concepts of differential and integral calculus—both critical in various fields of science, engineering, economics, and beyond. In calculus education, it is crucial to not only understand how to perform certain operations but also why they work and when they should be applied.

Problem-solving sessions that involve exercises, such as the one provided, help students solidify their understanding of integration by parts—a technique that can initially seem counterintuitive. By breaking down the problem and applying the technique step by step, students learn how integration by parts can unravel complex problems and reveal insightful connections between different mathematical expressions.

Improving Understanding in Calculus

To improve students' grasp of concepts like integration by parts, calculus education often utilizes visual aids, interactive elements, and real-world examples that demonstrate the practical applications of calculus. Moreover, encouraging students to work through several practice problems, especially those where the step-by-step solutions are thoroughly explained, prepares them to tackle a wide array of problems on their own and gives them a deeper understanding of the subject matter.