Problem 75 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Prove that if a function \(f\) has a continuous second derivative on an interval \([a, b]\) and \(f'(a) = f'(b) = 0\), then the following formula holds: \(\int_{a}^{b} x f^{\prime \prime}(x) d x=f(a)-f(b)\).

Step 1: Set up integration by parts

We have the integral we want to prove: $$ \int_{a}^{b} x f^{\prime \prime}(x) dx $$ To use integration by parts, we first choose which parts of the integral correspond to \(u\) and \(dv\). In this case, we choose: $$ u = x \quad \text{and} \quad dv = f^{\prime \prime}(x) dx $$ Now, we need to find the differential \(du\) and the function \(v\) such that \(dv = v' dx\). In our case: $$ du = dx \quad \text{and} \quad v = f'(x) $$ Now we can apply the integration by parts formula: $$ \int_{a}^{b} x f^{\prime \prime}(x) dx = \left[ xv - \int v du \right]_{a}^{b} $$ Which simplifies to: $$ \int_{a}^{b} x f^{\prime \prime}(x) dx = \left[ x f^{\prime}(x) - \int f^{\prime}(x) dx \right]_{a}^{b} $$

Step 2: Evaluate the expression

Using the Fundamental Theorem of Calculus, we can now evaluate the expression: $$ \int_{a}^{b} x f^{\prime \prime}(x) dx = \left[xf^{\prime}(x)\right]_a^b - \left[f^{\prime}(x)\right]_a^b = b f^{\prime}(b) - a f^{\prime}(a) - ( f^{\prime}(b) - f^{\prime}(a)) $$ But, given that \(f^{\prime}(a) = f^{\prime}(b) = 0\), the above expression reduces to: $$ \int_{a}^{b} x f^{\prime \prime}(x) dx = f(a) - f(b) $$ We have shown that if \(f\) has a continuous second derivative on \([a, b]\) and \(f^{\prime}(a)=f^{\prime}(b)=0\), then: $$ \int_{a}^{b} x f^{\prime \prime}(x) d x=f(a)-f(b) $$

Key Concepts

Continuous Second Derivative

In calculus, when we say a function has a continuous second derivative on an interval \( [a, b] \), it means that not only does the function itself exist and is continuous on this interval, but also that its first and second derivatives are continuous as well. Continuity of the second derivative, denoted as \( f''(x) \), is a sign of smoothness of the function's graph and implies the absence of sharp corners or cusps in that interval.

Why is this important? Because it ensures that the integration by parts, which relies on the derivative being well-behaved, is valid throughout the interval. For the integral \( \int_{a}^{b} x f''(x) dx \), the continuous nature of \( f''(x) \) on \( [a, b] \) serves as a guarantee for the legitimacy of operations we perform during integration, such as applying the Fundamental Theorem of Calculus. Furthermore, the conditions \( f'(a) = f'(b) = 0 \) hint at the function having horizontal tangents at the endpoints, which plays a key role in simplifying the final expression in our integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration—two of the main operations in calculus. This theorem comes in two parts, one of which (Part 1) states that if a function \( f \) is continuous on \( [a, b] \), then the function \( g(x) = \int_{a}^{x} f(t) dt \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \) and for every point \( x \) in that interval, \( g'(x) = f(x) \). The second part of the theorem (Part 2) tells us that if \( f \) is continuous on \( [a, b] \) and \( F \) is any antiderivative of \( f \) on that interval, then \( \int_{a}^{b} f(x) dx = F(b) - F(a) \).

In the context of our problem, we use the second part to simply calculate the antiderivative of \( f'(x) \) which is \( f(x) \) due to the properties of derivatives. Since \( f' \) is the derivative of \( f \) and is assumed to be continuous, we can directly evaluate the integral using the limits \( a \) and \( b \) to find the overall change in \( f \) on the interval.

Integration Techniques

Various integration techniques are available to solve more complex integrals. Integration by parts is one such technique that comes in handy when dealing with products of functions. It's based on the product rule for derivatives and is summarized by the formula \( \int u dv = uv - \int v du \), where \( u \) and \( v \) are functions of \( x \) that are chosen to simplify the integral.

In the given example, the choice \( u = x \) and \( dv = f''(x)dx \) is deliberate to reduce the integral to a more manageable form. The real art in applying integration by parts is choosing \( u \) and \( dv \) such that the resulting integral \( \int v du \) is easier to evaluate than the original.

Advice for Students

While practicing integration techniques, make sure to:

• Identify the patterns that suggest the use of a particular technique, like the product of functions for integration by parts.
• Understand the properties of the functions involved, such as continuity and differentiability over the given interval.
• Practice rearranging the integral if the first attempt at integration by parts doesn't simplify the problem.

Remember that mastering integration techniques requires practice and familiarity with a variety of functions and their antiderivatives.