Question

An identity Show that if \(f\) has continuous derivatives on \([a, b]\) and \(f^{\prime}(a)=f^{\prime}(b)=0,\) then $$\int_{a}^{b} x f^{\prime \prime}(x) d x=f(a)-f(b)$$.

Short answer

Question: Prove the identity $$\int_{a}^{b} x f^{\prime \prime}(x) d x=f(a)-f(b)$$ for a function \(f\) which has continuous second derivative and \(f'(a) = f'(b) = 0\). Answer: To prove this identity, we follow these steps: 1. Integrate the left-hand side by parts, using \(u = x\) and \(dv = f''(x) dx\). 2. Evaluate the new integral obtained. 3. Compare the result with the right-hand side. After following these steps, we have shown that the left-hand side and right-hand side of the given identity are equal, and therefore, we have proven the identity.

Step-by-step solution

Integrate by Parts

To integrate the left-hand side, we use integration by parts, with \(u = x\) and \(dv = f''(x) dx\). Then, we find \(du\) and \(v\) as follows: $$du = dx, \quad v = \int f''(x) dx = f'(x)$$ Now we apply the integration by parts formula: $$\int_{a}^{b} x f^{\prime \prime}(x) dx = uv \Big|_a^b - \int_{a}^{b} v du$$

Evaluate the New Integral

We now evaluate the right-hand side of the integration by parts equation: $$xf'(x) \Big|_a^b - \int_{a}^{b} f'(x) dx = (bf'(b) - af'(a)) - (F(b) - F(a))$$ Note that we use \(F(x)\) as the antiderivative of \(f'(x)\), which means \(F'(x) = f'(x)\). Also, \(f'(a) = f'(b) = 0\). We can then simplify the expression: $$= (b \cdot 0 - a \cdot 0) - (F(b) - F(a)) = -F(b) + F(a)$$

Compare the Result with the Right-Hand Side of the Original Identity

The right-hand side of the given identity is \(f(a) - f(b)\). Now, recall that \(F'(x) = f'(x)\). This means that \(F(x) = f(x) + C\) for some constant \(C\). Plugging this in, we get: $$-F(b) + F(a) = - (f(b) + C) + (f(a) + C) = f(a) - f(b)$$ This shows that the left-hand side and right-hand side of the given identity are equal, and therefore, we have proven the identity: $$\int_{a}^{b} x f^{\prime \prime}(x) dx = f(a) - f(b)$$

Key concepts

Calculus

Calculus is a branch of mathematics that focuses on rates of change and the accumulation of quantities. It's like the toolbox that helps us understand dynamic systems and solve complex problems. In particular, calculus is often divided into two primary segments:

• Differential Calculus: This is about finding rates of change. We use it to calculate derivatives, which tell us how a function changes at any point. Imagine you're driving a car; the derivative tells you your speed at any given moment.
• Integral Calculus: This part deals with accumulation, like areas under curves. It's the reverse process of differentiation, kind of like how you might think about the total distance you've driven over a trip.

Integration by parts, used in this exercise, is a technique from integral calculus named after the product rule for derivatives. It is crucial for integrating products of functions, especially when functions are not easily integrable in their original form.

Continuous Derivatives

Continuous derivatives tell us about the smoothness of a function. To say that a function's derivatives are continuous means there are no sudden jumps or breaks. This is crucial because continuous functions lead to more predictable and manageable behaviors. In the exercise, the function \(f\) has continuous derivatives on \([a, b]\). This means:

• \(f'\), \(f''\) (and so on) don’t have any gaps or abrupt changes on this interval.
• The function is smooth and well-behaved throughout the interval \([a, b]\).

These properties ensure we can apply calculus tools effectively. Specifically, knowing \(f'\) is continuous is critical because it allows us to integrate \(f''\), leading us to derive the result of the problem. Continuous derivatives also support the upcoming use of integral identities and makes sure boundary conditions (like \(f'(a) = f'(b) = 0\)) hold true without complications.

Integral Identities

Integral identities are powerful equations involving integrals, often simplifying the computation of complex integrals by expressing them in terms of known information. In this problem, the goal is to verify an integral identity:\[\int_{a}^{b} x f''(x) \, dx = f(a) - f(b)\]Using integration by parts confirmed this identity. Here's how it works step-by-step:

• Identify \(u\) and \(dv\). For this problem: \(u = x\) and \(dv = f''(x) \, dx\).
• Compute \(du\) and \(v\) using differentiation and integration: \(du = dx\), \(v = f'(x)\) since \(v = \int f''(x) \, dx\).
• Apply the integration by parts formula \(\int u \, dv = uv - \int v \, du\).
• The boundary terms evaluate to zero because \(f'(a) = f'(b) = 0\), simplifying the equation to one involving known values of \(f(x)\) only.

This identity offers an elegant way to connect the integral of derivatives to function values at boundaries. It's a testament to how integration, derivatives, and function values can combine to produce insightful results.