Question

If \(F\) and \(G\) are absolutely continuous on \([a, b]\), then so is \(F G\), and $$ \int_{a}^{b}\left(F G^{\prime}+G F^{\prime}\right)(x) d x=F(b) G(b)-F(a) G(a) $$

Short answer

The absolute continuity of \(F \) and \(G\) ensures \(F G\)' integrates to \(F(b)G(b) - F(a)G(a)\).

Step-by-step solution

Understanding Absolute Continuity

Two functions \( F \) and \( G \) are absolutely continuous on the interval \([a, b]\). This means that both functions can be expressed as integrals of their derivatives over any subinterval \([c, d] \subseteq [a, b]\). The properties of absolute continuity will be used to show that the product \( FG \) shares this characteristic.

Exploring the Product Rule

Recall that the derivative of the product of two functions is \((FG)' = F'G + FG'\). This will be pivotal in integrating the product \(FG\) since in absolute continuity, these derivative expressions can be manipulated to show that their respective integrals over \([a, b]\) also exhibit absolute continuity.

Applying Integration by Parts Conceptually

Rewrite \(FG\)' explicitly as \(F'G + FG'\) and acknowledge that since \(F \) and \( G\) are absolutely continuous, we can find functions \(p\) and \(q\) such that \(F(x) = F(a) + \int_a^x p(t) dt\) and \(G(x) = G(a) + \int_a^x q(t) dt\). These integrals exist over any subinterval, thus \(F' = p\) and \(G' = q\) are their derivatives respectively.

Deriving from Integration by Parts Formula

Use the integration by parts for the integral \( \int_a^b (F G)'(x) \, dx \). Substituting \((FG)' = F'G + FG'\) into the integral, we get:\[\int_a^b (FG)'(x) \, dx = \int_a^b (F'G + FG')(x) \, dx.\]This integral evaluates to \(F(b)G(b) - F(a)G(a)\) due to the Fundamental Theorem of Calculus.

Confirming the Integration Result

The integral \( \int_a^b (F G' + G F')(x) \, dx \) matches the integral of the derivative of the product \((FG)'\), confirming the evaluation to be \[F(b)G(b) - F(a)G(a)\]. Hence, the expression \(F G' + G F'\) gives the change in the product \(FG\) over \([a, b]\), proving the formula using absolute continuity properties.

Key concepts

Product Rule

The product rule is an essential concept in calculus, especially when dealing with the derivatives of functions that are products of two variables. The product rule states that if you have two differentiable functions, say \( F(x) \) and \( G(x) \), then their product \( FG(x) \) has a derivative given by:

• \((FG)' = F'G + FG'\)

This rule is a crucial tool because it allows us to find the rate of change of the product of two functions without independently differentiating them first.

In the context of absolute continuity, as seen in our exercise, the product rule reveals how to handle the expressions involving derivatives like \( F'(x)G(x) + F(x)G'(x) \). This is important when both \( F \) and \( G \) are absolutely continuous functions defined over the interval \([a, b]\). Using the product rule helps us break down complex expressions into manageable derivatives and demonstrates that \( FG \) itself is absolutely continuous.

Integration by Parts

Integration by parts is a method used to integrate products of two functions. It is particularly useful when direct integration is challenging. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

This formula is derived from the product rule of differentiation and is aimed at simplifying integrals of products into simpler components.

In the exercise, where we look at an integral involving \( F'G + FG' \), we apply integration by parts to explore and transform these terms. By setting one function as \( u \) and its derivative as \( du \), and another function as \( v \) with its derivative \( dv \), we can strategically simplify complicated integrations.

This technique not only makes evaluating the integral more manageable but also lays down the groundwork for proving that for absolutely continuous functions, the integral of their derivatives results in a straightforward expression, as shown in the solution.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful principle that bridges the gap between differentiation and integration, showing their inverse nature. It is composed of two parts:

• Part 1: It states that if \( f \) is a continuous real-valued function on an interval \([a, b]\) and \( F \) is an antiderivative of \( f \) on \([a, b]\), then:
  • \( \int_a^b f(x) \, dx = F(b) - F(a) \)
• Part 2: It states that if \( F \) is an antiderivative of \( f \) in an interval \([a, b]\):
  • \( \frac{d}{dx} \int_a^x f(t) \, dt = f(x) \)

This theorem is central to solving integrals, as it allows us to compute definite integrals using antiderivatives.

In our exercise context, the Fundamental Theorem of Calculus helps confirm that the evaluation of the integral \( \int_a^b (F'G + FG')(x) \, dx \) results in \( F(b)G(b) - F(a)G(a) \). This statement confirms the interplay between the function and its derivative, effectively showing changes over an interval. The theorem's role is crucial in validating the integration process and solving problems related to absolutely continuous functions.