Problem 21 - A Course in Calculus and Real Analysis

Question

Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be such that \(f\) is differentiable, \(f^{\prime}\) is integrable, and \(g\) is continuous. If $$ G(x):=\int_{a}^{x} g(t) d t \quad \text { and } \quad \tilde{G}(x):=\int_{x}^{b} g(t) d t \quad \text { for } x \in[a, b] $$ then show that $$ \int_{a}^{b} f(x) g(x) d x=f(b) G(b)-\int_{a}^{b} f^{\prime}(x) G(x) d x=f(a) \widetilde{G}(a)+\int_{a}^{b} f^{\prime}(x) \widetilde{G}(x) d x $$ (Compare Proposition 6.28.)

Step-by-Step Solution

Answer

In conclusion, we have shown that: \[ \int_{a}^{b} f(x) g(x) d x = f(b) G(b) - \int_{a}^{b} f^{\prime}(x) G(x) d x = f(a) \widetilde{G}(a) + \int_{a}^{b} f^{\prime}(x) \widetilde{G}(x) d x \]

Step 1: Define functions and their derivatives

Let u(x) = f(x) and du(x) = f^{\prime}(x) for both parts of the problem. Applying integration by parts, we compute:

Step 2: Apply integration by parts with G(x)

Let v(x) = G(x) = \(\int_{a}^{x} g(t) dt\) and dv(x) = g(x) for the first part. Now, use the formula for integration by parts: \[ \int_{a}^{b} f(x)g(x) dx = [f(x)G(x)]_{a}^{b} - \int_{a}^{b} f^{\prime}(x) G(x) dx \] So after simplifying, we obtain one part of the given equation: \[ \int_{a}^{b} f(x) g(x) d x = f(b) G(b) - \int_{a}^{b} f^{\prime}(x) G(x) d x \]

Step 3: Apply integration by parts with \(\widetilde{G}(x)\)

We will now let v(x) = \(\widetilde{G}(x)\) = \(\int_{x}^{b} g(t) dt\) and dv(x) = -g(x) for the second part. Applying the integration by parts formula again: \[ \int_{a}^{b} f(x)g(x) dx = -[f(x)\widetilde{G}(x)]_{a}^{b} + \int_{a}^{b} f^{\prime}(x) \widetilde{G}(x) dx \] After simplifying, we obtain the second part of the given equation: \[ \int_{a}^{b} f(x) g(x) d x = f(a) \widetilde{G}(a) + \int_{a}^{b} f^{\prime}(x) \widetilde{G}(x) d x \]

Step 4: Combine the results

In conclusion, we have shown that: \[ \int_{a}^{b} f(x) g(x) d x = f(b) G(b) - \int_{a}^{b} f^{\prime}(x) G(x) d x = f(a) \widetilde{G}(a) + \int_{a}^{b} f^{\prime}(x) \widetilde{G}(x) d x \]

Key Concepts

Differentiable Functions

Understanding differentiable functions is key when navigating through calculus problems, specifically when discussing integration by parts. A function is said to be differentiable over an interval, such as \[a, b\], if it has a derivative at every point within that interval. In simpler terms, if you can draw a tangent line to the function's curve at any point in that interval, then the function is differentiable there.

The significance of being differentiable is that it ensures the existence of the derivative function, commonly denoted as \(f'(x)\), which represents the rate of change of the function with respect to \(x\). This derivative function must also be integrable, meaning we can find the area under the curve represented by \(f'(x)\) over the interval \[a, b\].

In the context of the given exercise, the function \(f\) being differentiable and its derivative \(f'\) being integrable are crucial conditions for the application of integration by parts, which enables the solution of complex integral expressions by breaking them down into simpler parts.

Continuous Functions

Continuity in functions is another fundamental concept in integral calculus, interplaying with differentiability. A function \(g\) is continuous on an interval, such as \[a, b\], when you can draw its graph without lifting your pen from the paper. This implies that the function doesn’t have any gaps, jumps, or points of discontinuity in that interval.

The role of continuity in calculus is indispensable, as it guarantees the function behaves well and that limits and function values match at every point within said interval. It’s important to note that differentiable functions are always continuous, but continuous functions are not necessarily differentiable everywhere. For instance, a continuous function might have a sharp corner or cusp where it is not differentiable.

In our textbook problem, the continuous nature of \(g\) ensures that the function \(G(x)\), which is defined as the integral of \(g\) over \[a, x\], is well-behaved. This continuity of \(g\) is essential when integrating by parts, as it supports the existence of the integrals \(G(x)\) and \(\tilde{G}(x)\), which are pivotal in reaching the eventual solution of the integral equation.

Integral Calculus

Integral calculus, one of the two major branches of calculus, is concerned with the accumulation of quantities, such as areas under curves, amongst other applications. Integrals themselves can be classified into two kinds: definite and indefinite. In our exercise, we deal with a definite integral, which is an integral with upper and lower limits that provides a numerical value corresponding to the accumulated quantity.

The method of integration by parts is an essential tool in solving integrals that involve products of functions, essentially derived from the product rule for differentiation. It allows us to transform a complicated integral into simpler parts that can be more easily managed.

The formula for integration by parts is given by \[\int u dv = uv - \int v du\], where \(u\) and \(dv\) are functions of \(x\). This technique requires you to judiciously choose which function in the integrand to assign to \(u\) and which to differentiate to \(dv\), to simplify the integral-solving process.

In the context of our exercise, integration by parts enables the translation of the integral of the product of functions, \(f(x)g(x)\), into terms involving the integral of \(f'(x)\) multiplied by antiderivatives of \(g(x)\), illustrated as \(G(x)\) and \(\tilde{G}(x)\), thus leading to the detailed solution provided.