Question

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

Short answer

Question: Show that if \(f\) and \(g\) are functions with continuous second derivatives on the interval [0, 1] and they satisfy the boundary conditions \(f(0) = f(1) = g(0) = g(1) = 0\), then the identity \(\int_{0}^{1} f^{\prime \prime}(x) g(x) dx=\int_{0}^{1} f(x) g^{\prime \prime}(x) dx\) holds. Answer: Using integration by parts and applying the given boundary conditions, we find that \(\int_{0}^{1} f^{\prime \prime}(x) g(x) dx=\int_{0}^{1} f(x) g^{\prime \prime}(x) dx\).

Step-by-step solution

Integration by Parts Formula

The formula for integration by parts is given as: $$\int u \, dv = uv - \int v\, du$$ Where \(u\) and \(v\) are differentiable functions, and \(du\) and \(dv\) are their respective differentials.

Apply Integration by Parts

Let's apply the integration by parts formula to the first integral: $$\int_{0}^{1} f^{\prime \prime}(x) g(x) dx$$ Choose \(u = g(x)\) and \(dv = f^{\prime \prime}(x) dx\). Then, we have: $$du = g'(x) dx$$ $$v = \int f^{\prime \prime}(x) dx = f'(x) + C_1(x)$$ Now applying the integration by parts formula: $$\int_{0}^{1} f^{\prime \prime}(x) g(x) dx = \left[f'(x)g(x) + C_1(x)g(x)\right]_0^1 - \int_{0}^{1} (f'(x) + C_1(x)) g'(x) dx$$

Evaluate Boundary Values

Now we evaluate the first term at the boundaries: $$\left[f'(x)g(x) + C_1(x)g(x)\right]_0^1 = [f'(1)g(1) + C_1(1)g(1)] - [f'(0)g(0) + C_1(0)g(0)]$$ Because of the given boundary conditions, \(f(1)=0\), \(g(0)=0\), and \(g(1)=0\), we have: $$\left[f'(x)g(x) + C_1(x)g(x)\right]_0^1 = 0$$

Simplify Integral

We can now simplify the first integral: $$\int_{0}^{1} f^{\prime \prime}(x) g(x) dx = -\int_{0}^{1} (f'(x) + C_1(x)) g'(x) dx$$

Apply Integration by Parts

Now apply the integration by parts to the second integral: $$\int_{0}^{1} f(x) g^{\prime \prime}(x) dx$$ Choose \(u = f(x)\) and \(dv = g^{\prime \prime}(x) dx\). Then, we have: $$du = f'(x) dx$$ $$v = \int g^{\prime \prime}(x) dx = g'(x) + C_2(x)$$ Applying the integration by parts formula: $$\int_{0}^{1} f(x) g^{\prime \prime}(x) dx = \left[f(x)(g'(x) + C_2(x))\right]_0^1 - \int_{0}^{1} (f'(x) + C_2(x))g'(x)dx$$

Evaluate Boundary Values

Now we evaluate the first term at the boundaries: $$\left[f(x)(g'(x) + C_2(x))\right]_0^1 = [f(1)(g'(1) + C_2(1))] - [f(0)(g'(0) + C_2(0))]$$ Because of the given boundary conditions, \(f(0)=0\) and \(f(1)=0\), we have: $$\left[f(x)(g'(x) + C_2(x))\right]_0^1 = 0$$

Simplify Integral

We can now simplify the second integral: $$\int_{0}^{1} f(x) g^{\prime \prime}(x) dx = -\int_{0}^{1} (f'(x) + C_2(x))g'(x)dx$$

Compare Integrals

Now we compare the two simplified integrals: $$-\int_{0}^{1} (f'(x) + C_1(x)) g'(x) dx = -\int_{0}^{1} (f'(x) + C_2(x))g'(x)dx$$ Since the only difference between the two integrals is the constant term, we can conclude that the identity holds: $$\int_{0}^{1} f^{\prime \prime}(x) g(x) dx=\int_{0}^{1} f(x) g^{\prime \prime}(x) dx$$

Key concepts

Continuous Second Derivatives

Understanding continuous second derivatives is crucial when solving complex problems in calculus. When a function has continuous second derivatives, it means that not only is the function itself continuous, but its first and second derivatives are also continuous across their domain. Continuity of these derivatives ensures the absence of 'sharp' turns or 'breaks' in the graph of the derivatives. This property plays a vital role in certain calculus techniques, such as integration by parts, where taking derivatives is a fundamental step.

In our context, given that both functions f and g possess continuous second derivatives, we are guaranteed that the integrals of these functions will be well defined and devoid of unexpected behavior, which is often caused by discontinuities. This property simplifies the use of integration by parts, as we can reliably integrate and differentiate these functions across the interval from 0 to 1.

Integral Calculus

The realm of integral calculus explores the concept of finding the area under curves, antiderivatives, and solving problems involving accumulation. One technique in integral calculus is integration by parts, which is derived from the product rule for differentiation and is used to integrate the product of two functions.

Application of Integration by Parts

The key to solving the given exercise is applying the integration by parts formula cleverly, so that the boundary conditions can be used effectively. By selecting appropriate functions for u and dv, the integration by parts formula allows us to transform complex integrals into simpler forms. Once transformed, it becomes apparent that the boundary conditions impose that certain terms vanish, leaving behind integrals that can be directly compared. Consequently, this demonstrates that the initial identity holds, showcasing the power of integration by parts in integral calculus.

Boundary Conditions

Boundary conditions are essential when solving differential equations and evaluating integrals, as they provide additional information about the solutions at the boundaries of the domain. They effectively 'anchor' the solution at specific points, which allows for the unique determination of constants that arise when integrating.

In the solved problem, the boundary conditions state that f(0)=f(1)=g(0)=g(1)=0. These conditions have substantial implications during the evaluation of the integrals after applying integration by parts. The resulting terms involving f and g and their derivatives at the points 0 and 1 are set to zero due to these conditions. The simplification leads to the conclusion that these terms do not contribute to the final value of the integral, which is crucial in proving the identity. Thus, understanding how to apply boundary conditions is invaluable in confirming the equality of the two integrals.